NCERT Solutions Class 11 Maths Chapter 11

Q.1 In each of the following Exercises 1 to 5, find the equation of the circle with
1. centre (0,2) and radius 2
2. centre (–2,3) and radius 4
3.

$\mathbf{centre}\mathbf{(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{)}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\frac{\mathbf{1}}{\mathbf{12}}$

4.

$\mathbf{centre}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\sqrt{\mathbf{2}}$

5.

$\mathbf{centre}\mathbf{(}\mathbf{-}\mathbf{a}\mathbf{,}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}$

Ans

1.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\\ \mathrm{Equation}\text{of circle with centre}(0,2)\text{and radius 2 is:}\\ {(\mathrm{x}-0)}^2+{(\mathrm{y}-2)}^2=2^2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{y}+4=4\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{y}=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

2.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\\ \mathrm{Equation}\text{of circle with centre}(-2,3)\text{and radius 4 is:}\\ {(\mathrm{x}+2)}^2+{(\mathrm{y}-3)}^2=4^2\\ \Rightarrow {\mathrm{x}}^2+4\mathrm{x}+4+{\mathrm{y}}^2-6\mathrm{y}+9=16\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}-3=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

3.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\\ \mathrm{Equation}\text{of circle with centre}(\frac{1}{2},\frac{1}{4})\text{and radius}\frac{1}{12}\text{is:}\\ {(\mathrm{x}-\frac{1}{2})}^2+{(\mathrm{y}-\frac{1}{4})}^2={\left(\frac{1}{12}\right)}^2\\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}+\frac{1}{4}+{\mathrm{y}}^2-\frac{1}{2}\mathrm{y}+\frac{1}{16}=\frac{1}{144}\\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}+{\mathrm{y}}^2-\frac{1}{2}\mathrm{y}+\frac{1}{4}+\frac{1}{16}-\frac{1}{144}=0\\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}+{\mathrm{y}}^2-\frac{1}{2}\mathrm{y}+\frac{11}{36}=0\\ \Rightarrow 36{\mathrm{x}}^2+36{\mathrm{y}}^2-36\mathrm{x}-18\mathrm{y}+11=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

4.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\\ \mathrm{Equation}\text{of circle with centre}(1,1)\text{and radius}\sqrt{\text{2}}\text{is:}\end{array}$ \begin{array}{}\end{array} $\begin{array}{l}{(\mathrm{x}-1)}^2+{(\mathrm{y}-1)}^2={\left(\sqrt{\text{2}}\right)}^2\\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}+1+{\mathrm{y}}^2-2\mathrm{y}+1=2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{y}=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

5.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\\ \mathrm{Equation}\text{of circle with centre}(-\mathrm{a},-\mathrm{b})\text{and radius}\sqrt{{\text{a}}^{\text{2}}-{\mathrm{b}}^2}\text{is:}\\ {(\mathrm{x}+\mathrm{a})}^2+{(\mathrm{y}+\mathrm{b})}^2={\left(\sqrt{{\text{a}}^{\text{2}}-{\mathrm{b}}^2}\right)}^2\\ {\mathrm{x}}^2+2\mathrm{ax}+{\mathrm{a}}^2+{\mathrm{y}}^2+2\mathrm{by}+{\mathrm{b}}^2={\text{a}}^{\text{2}}-{\mathrm{b}}^2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{ax}+2\mathrm{by}+2{\mathrm{b}}^2=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.2 In each of the following Exercises 6 to 9, find the centre and radius of the circles.

6. (x + 5)² + (y – 3)² = 36
7. x² + y² – 4x – 8y – 45 = 0
8. x² + y² – 8x + 10y – 12 = 0
9. 2x² + 2y² – x = 0

Ans

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\text{(x+5)}}^{\text{2}}\text{+(y}-{\text{3)}}^{\text{2}}\text{=36}...\left(\text{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\text{and we get}-\\ \mathrm{h}=-5,\mathrm{k}=3{\text{and r}}^{\text{2}}=\text{36}\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{h}=-5,\mathrm{k}=3\text{and r}=\text{6}\\ \text{Therefore, centre is}(-5,3)\text{and radius is 6.}\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-8\mathrm{y}-45=0\\ \Rightarrow ({\mathrm{x}}^2-4\mathrm{x})+({\mathrm{y}}^2-8\mathrm{y})-45=0\\ \Rightarrow ({\mathrm{x}}^2-4\mathrm{x}+2^2-2^2)+({\mathrm{y}}^2-8\mathrm{y}+4^2-4^2)-45=0\\ \Rightarrow ({\mathrm{x}}^2-4\mathrm{x}+2^2)-4+({\mathrm{y}}^2-8\mathrm{y}+4^2)-16-45=0\\ \Rightarrow {({\mathrm{x}}^2-2)}^2+{({\mathrm{y}}^2-4)}^2-65=0\\ \Rightarrow {({\mathrm{x}}^2-2)}^2+{({\mathrm{y}}^2-4)}^2={\left(\sqrt{65}\right)}^2...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\text{and we get}-\\ \mathrm{h}=2,\mathrm{k}=4\text{and r}=\sqrt{65}\\ \text{Therefore, centre is}(2,4)\text{and radius is}\sqrt{\text{65}}\text{.}\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}+10\mathrm{y}-12=0\\ \Rightarrow ({\mathrm{x}}^2-8\mathrm{x})+({\mathrm{y}}^2+10\mathrm{y})-12=0\\ \Rightarrow ({\mathrm{x}}^2-8\mathrm{x}+4^2-4^2)+({\mathrm{y}}^2+10\mathrm{y}+5^2-5^2)-12=0\\ \Rightarrow ({\mathrm{x}}^2-8\mathrm{x}+4^2)-16+({\mathrm{y}}^2+10\mathrm{y}+5^2)-25-12=0\end{array}$ $\begin{array}{l}\Rightarrow {({\mathrm{x}}^2-4)}^2+{({\mathrm{y}}^2+5)}^2-53=0\\ \Rightarrow {({\mathrm{x}}^2-4)}^2+{({\mathrm{y}}^2+5)}^2={\left(\sqrt{53}\right)}^2...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\text{and we get}-\\ \mathrm{h}=4,\mathrm{k}=-5\text{and r}=\sqrt{53}\\ \text{Therefore, centre is}(4,-5)\text{and radius is}\sqrt{\text{53}}\text{.}\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}2x}}^{\text{2}}+{\text{2y}}^{\text{2}}-\text{x}=0\\ \Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}x}}^{\text{2}}+{\text{y}}^{\text{2}}-\frac{1}{\text{2}}\text{x}=0\\ \Rightarrow ({\mathrm{x}}^2-\frac{1}{2}\mathrm{x})+{\mathrm{y}}^2=0\\ \Rightarrow ({\mathrm{x}}^2-\frac{1}{2}\mathrm{x}+\frac{1}{4^2}-\frac{1}{4^2})+{\mathrm{y}}^2=0\\ \Rightarrow ({\mathrm{x}}^2-\frac{1}{2}\mathrm{x}+\frac{1}{4^2})+{\mathrm{y}}^2=\frac{1}{2^2}\\ \Rightarrow {(\mathrm{x}-\frac{1}{4})}^2+{\mathrm{y}}^2=\frac{1}{4^2}...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2\text{and we get}-\\ \mathrm{h}=\frac{1}{4},\mathrm{k}=0\text{and r}=\frac{1}{4}\end{array}$ $\text{Therefore, centre is}(\frac{1}{4},0)\text{and radius is}\frac{1}{4}\text{.}$

Q.3 Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

Ans

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{i}\right)\\ \mathrm{Circle}\left(\mathrm{i}\right)\text{is passing through}(4,1)\text{and}(6,5)\text{then,}\\ {(4-\mathrm{h})}^2+{(1-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{ii}\right)\\ \mathrm{and}{(6-\mathrm{h})}^2+{(5-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{\hspace{0.17em}and equation}\left(\mathrm{iii}\right),\text{we get}\\ {(4-\mathrm{h})}^2+{(1-\mathrm{k})}^2={(6-\mathrm{h})}^2+{(5-\mathrm{k})}^2\\ \Rightarrow 16-8\mathrm{h}+{\mathrm{h}}^2+1-2\mathrm{k}+{\mathrm{k}}^2=36-12\mathrm{h}+{\mathrm{h}}^2+25-10\mathrm{k}+{\mathrm{k}}^2\\ \Rightarrow 4\mathrm{h}+8\mathrm{k}=44\\ \Rightarrow \mathrm{h}+2\mathrm{k}=11...\left(\mathrm{iv}\right)\\ \mathrm{Since},\text{centre}(\mathrm{h},\mathrm{k})\text{lies on the line 4x}+\mathrm{y}=16.\\ \mathrm{So},4\mathrm{h}+\mathrm{k}=16...\left(\mathrm{v}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{iv}\right)\text{and equation}\left(\mathrm{v}\right),\text{we get}\\ \text{h}=\text{3 and k}=\text{4}\\ \text{Putting value of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ {(4-3)}^2+{(1-4)}^2={\mathrm{r}}^2\\ \Rightarrow {\mathrm{r}}^2=1+9\\ =10\end{array}$ $\begin{array}{l}\mathrm{Then},\text{equation of circle is:}\\ {(\mathrm{x}-3)}^2+{(\mathrm{y}-4)}^2=10\\ \Rightarrow {\mathrm{x}}^2-6\mathrm{x}+9+{\mathrm{y}}^2-8\mathrm{y}+16=10\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-8\mathrm{y}+15=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.4 Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.

Ans

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{i}\right)\\ \mathrm{Circle}\left(\mathrm{i}\right)\text{is passing through}(2,3)\text{and}(-1,1)\text{then,}\\ {(2-\mathrm{h})}^2+{(3-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{ii}\right)\\ \mathrm{and}{(-1-\mathrm{h})}^2+{(1-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{\hspace{0.17em}and equation}\left(\mathrm{iii}\right),\text{we get}\\ {(2-\mathrm{h})}^2+{(3-\mathrm{k})}^2={(-1-\mathrm{h})}^2+{(1-\mathrm{k})}^2\\ \Rightarrow 4-4\mathrm{h}+{\mathrm{h}}^2+9-6\mathrm{k}+{\mathrm{k}}^2=1+2\mathrm{h}+{\mathrm{h}}^2+1-2\mathrm{k}+{\mathrm{k}}^2\\ \Rightarrow -6\mathrm{h}-4\mathrm{k}=-11\\ \Rightarrow 6\mathrm{h}+4\mathrm{k}=11...\left(\mathrm{iv}\right)\\ \mathrm{Since},\text{centre}(\mathrm{h},\mathrm{k})\text{lies on the line x}-3\mathrm{y}=11.\\ \mathrm{So},\mathrm{h}-3\mathrm{k}=11...\left(\mathrm{v}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{iv}\right)\text{and equation}\left(\mathrm{v}\right),\text{we get}\\ \text{h}=\frac{7}{2}\text{and k}=-\frac{\text{5}}{2}\end{array}$ $\begin{array}{l}\text{Putting value of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ {(2-\frac{7}{2})}^2+{(3+\frac{\text{5}}{2})}^2={\mathrm{r}}^2\\ \Rightarrow {\mathrm{r}}^2=\frac{9}{4}+\frac{121}{4}\\ =\frac{130}{4}\\ \mathrm{Then},\text{equation of circle is:}\\ {(\mathrm{x}-\frac{7}{2})}^2+{(\mathrm{y}-\frac{\text{5}}{2})}^2=\frac{130}{4}\\ \Rightarrow {\mathrm{x}}^2-7\mathrm{x}+\frac{49}{4}+{\mathrm{y}}^2+5\mathrm{y}+\frac{25}{4}=\frac{130}{4}\\ \Rightarrow 4{\mathrm{x}}^2+4{\mathrm{y}}^2-28\mathrm{x}+20\mathrm{y}=56\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-7\mathrm{x}+5\mathrm{y}-14=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.5 Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

Ans

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{a},0)\text{and radius 5 is:}\\ {(\mathrm{x}-\mathrm{a})}^2+{(\mathrm{y}-0)}^2=5^2\\ \Rightarrow {(\mathrm{x}-\mathrm{a})}^2+{\mathrm{y}}^2=5^2...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{circle}\left(\mathrm{i}\right)\text{passes through the point}(2,3).\mathrm{So},\\ {(2-\mathrm{a})}^2+3^2=5^2\\ \Rightarrow 4-4\mathrm{a}+{\mathrm{a}}^2+9=25\\ \Rightarrow {\mathrm{a}}^2-4\mathrm{a}=25-13\end{array}$ $\begin{array}{l}\Rightarrow {\mathrm{a}}^2-4\mathrm{a}-12=0\\ \Rightarrow (\mathrm{a}-6)(\mathrm{a}+2)=0\\ \Rightarrow \mathrm{a}=6,-2\\ \mathrm{Case}1:\mathrm{When}\text{centre of circle is}(6,0):\\ \mathrm{The}\text{equation of circle is:-}\\ {(\mathrm{x}-6)}^2+{(\mathrm{y}-0)}^2=5^2\\ \Rightarrow {\mathrm{x}}^2-12\mathrm{x}+36+{\mathrm{y}}^2=25\\ \Rightarrow {\mathrm{x}}^2-12\mathrm{x}+{\mathrm{y}}^2+11=0\\ \mathrm{Case}2:\mathrm{When}\text{centre of circle is}(-2,0):\\ \mathrm{The}\text{equation of circle is:-}\\ {(\mathrm{x}+2)}^2+{(\mathrm{y}-0)}^2=5^2\\ \Rightarrow {\mathrm{x}}^2+4\mathrm{x}+4+{\mathrm{y}}^2=25\\ \Rightarrow {\mathrm{x}}^2+4\mathrm{x}+{\mathrm{y}}^2-21=0\\ \mathrm{Thus},\text{the equation of circle is}\\ {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-21=0\text{or}{\mathrm{x}}^2+{\mathrm{y}}^2-12\mathrm{x}+11=0.\end{array}$

Q.6 Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Ans

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(\mathrm{h},\mathrm{k})\text{and radius r is:}\\ {(\mathrm{x}-\mathrm{h})}^2+{(\mathrm{y}-\mathrm{k})}^2={\mathrm{r}}^2...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{passes through the points}(0,0),(\mathrm{a},0)\text{and}(0,\mathrm{b}),\text{so}\\ {(0-\mathrm{h})}^2+{(0-\mathrm{k})}^2={\mathrm{r}}^2\\ \Rightarrow {\mathrm{h}}^2+{\mathrm{k}}^2={\mathrm{r}}^2...\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}{(\mathrm{a}-\mathrm{h})}^2+{(0-\mathrm{k})}^2={\mathrm{r}}^2\\ \Rightarrow {\mathrm{a}}^2-2\mathrm{ah}+{\mathrm{h}}^2+{\mathrm{k}}^2={\mathrm{r}}^2...\left(\mathrm{iii}\right)\\ {(0-\mathrm{h})}^2+{(\mathrm{b}-\mathrm{k})}^2={\mathrm{r}}^2\\ \Rightarrow {\mathrm{h}}^2+{\mathrm{b}}^2-2\mathrm{bk}+{\mathrm{k}}^2={\mathrm{r}}^2...\left(\mathrm{iv}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we get}\\ {\mathrm{h}}^2+{\mathrm{k}}^2={\mathrm{a}}^2-2\mathrm{ah}+{\mathrm{h}}^2+{\mathrm{k}}^2\\ \Rightarrow {\mathrm{a}}^2-2\mathrm{ah}=0\\ \Rightarrow \mathrm{a}(\mathrm{a}-2\mathrm{h})=0\\ \Rightarrow \mathrm{a}-2\mathrm{h}=0,\mathrm{a}\ne 0\\ \Rightarrow \mathrm{h}=\frac{\mathrm{a}}{2}\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iv}\right),\text{we get}\\ {\mathrm{h}}^2+{\mathrm{k}}^2={\mathrm{h}}^2+{\mathrm{b}}^2-2\mathrm{bk}+{\mathrm{k}}^2\\ \Rightarrow {\mathrm{b}}^2-2\mathrm{bk}=0\\ \Rightarrow \mathrm{b}(\mathrm{b}-2\mathrm{k})=0\\ \Rightarrow \mathrm{b}-2\mathrm{k}=0,\mathrm{b}\ne 0\\ \Rightarrow \mathrm{k}=\frac{\mathrm{b}}{2}\\ \mathrm{Putting}\text{values of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ {\left(\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{b}}{2}\right)}^2={\mathrm{r}}^2\\ \frac{{\mathrm{a}}^2}{4}+\frac{{\mathrm{b}}^2}{4}={\mathrm{r}}^2\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right):\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}{(\mathrm{x}-\frac{\mathrm{a}}{2})}^2+{(\mathrm{y}-\frac{\mathrm{b}}{2})}^2=\frac{{\mathrm{a}}^2}{4}+\frac{{\mathrm{b}}^2}{4}\\ \Rightarrow {\mathrm{x}}^2-\mathrm{ax}+\frac{{\mathrm{a}}^2}{4}+{\mathrm{y}}^2-\mathrm{by}+\frac{{\mathrm{b}}^2}{4}=\frac{{\mathrm{a}}^2}{4}+\frac{{\mathrm{b}}^2}{4}\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{ax}-\mathrm{by}=0\\ \mathrm{This}\text{is the required equation of the circle.}\end{array}$

Q.7 Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Ans

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}(2,2)\text{and radius r is:}\\ {(\mathrm{x}-2)}^2+{(\mathrm{y}-2)}^2={\mathrm{r}}^2...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{passes through the point}(4,5),\text{so}\\ {(4-2)}^2+{(5-2)}^2={\mathrm{r}}^2\\ \Rightarrow 4+9={\mathrm{r}}^2\\ \Rightarrow 13={\mathrm{r}}^2\\ \mathrm{Putting}{\text{value of r}}^{\text{2}}\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ {(\mathrm{x}-2)}^2+{(\mathrm{y}-2)}^2=13\\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}+4+{\mathrm{y}}^2-4\mathrm{y}+4=13\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{y}=5\\ \mathrm{This}\text{is the equation of required circle.}\end{array}$

Q.8 Does the point (–2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?

Ans

$\begin{array}{l}\mathrm{The}\text{equation of given circle is:}\\ {\text{x}}^{\text{2}}+{\text{y}}^{\text{2}}=\text{25}\end{array}$ $\begin{array}{l}\text{Putting x}=\text{2}.\text{5},\text{y}=\text{3}.\text{5 in L.H.S. of given equation, we get}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{2}}+{\text{y}}^{\text{2}}={(\text{2}.\text{5})}^{\text{2}}+{(\text{3}.\text{5})}^{\text{2}}\\ =6.25+12.25\\ =18.50<25=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{point}(–\text{2}.\text{5},\text{3}.\text{5})\text{lies inside of the circle.}\\ \text{As distance of the point from centre is less than radius}\\ \text{of the circle.}\end{array}$

Q.9 In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

1. y² = 12x
2. x² = 6y
3. y² = – 8x
4. x² = – 16y
5. y² = 10x
6. x² = – 9y

Ans

1.

The given equation of parabola is:
y² = 12x
Since, coefficient of x is positive. So, comparing it with y² = 4ax and we get
4a = 12
a = 3
The coordinates of the focus is (3, 0).
Since, this equation has y², so axis is x-axis.
Equation of directrix is: x = – 3.
Length of latus rectum = 4a
= 4 x 3
= 12

2.

The given equation of parabola is: x² = 6y
Since, coefficient of y is positive. So, comparing it with x² = 4ay and we get
4a = 6
a = (3/2)
The coordinates of the focus is (0, 3/2).
Since, this equation has x², so axis is y-axis.
Equation of directrix is: y = – 3/2.
Length of latus rectum = 4a
= 4 x 3/2
= 6

3.

The given equation of parabola is:
y² = – 8x
Since, coefficient of x is negative. So, comparing it with y² = – 4ax and we get
– 4a = – 8
a = 2
The coordinates of the focus is (– 2, 0).
Since, this equation has y², so axis is x-axis.
Equation of directrix is:
x = 2.
Length of latus rectum = 4a
= 4 x 2
= 8

4.

The given equation of parabola is:
x² = – 16y
Since, coefficient of y is negative. So, comparing it with x² = – 4ay and we get
4a = – 16
a = –4
The coordinates of the focus is (0, –4).
Since, this equation has x², so axis is y-axis.
Equation of directrix is: y = 4.
Length of latus rectum = 4a
= 4 x 4
= 16

5.

The given equation of parabola is:
y² = 10x
Since, coefficient of x is positive. So, comparing it with y² = 4ax and we get
4a = 10
a = (5/2)
The coordinates of the focus is (5/2, 0).
Since, this equation has y², so axis is x-axis.
Equation of directrix is: x = – 5/2.
Length of latus rectum = 4a
= 4 x 5/2
= 10

6.

The given equation of parabola is:
x² = – 9y
Since, coefficient of y is negative. So, comparing it with x² = – 4ay and we get
4a = – 9
a = – 9/4
The coordinates of the focus is (0, – 9/4).
Since, this equation has x², so axis is y-axis.
Equation of directrix is: y = 9/4.
Length of latus rectum = 4a
= 4 x 9/4
= 9

Q.10 In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
7. Focus (6,0); directrix x = – 6
8. Focus (0,–3); directrix y = 3
9. Vertex (0,0); focus (3,0)
10. Vertex (0,0); focus (–2,0)
11. Vertex (0,0) passing through (2,3) and axis is along x-axis.
12. Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.

Ans

7.

Since, focus (6, 0) lies on x-axis and directrix
x = – 6 shows that parabola is in the form of
y² = 4ax where a = 6.
Therefore, the equation of parabola is:
y² = 4(6)x
y² = 24 x

8.

Since, focus (0,–3) lies on y-axis and directrix
y = 3 shows that parabola is in the form of
x² = – 4ay where a = 3.
Therefore, the equation of parabola is:
x² = – 4(3)y
x² = – 12y

9.

Since, vertex is (0, 0) and focus is (3, 0) which lies on x-axis of parabola.
Therefore, the equation of parabola will be in the form of y² = 4ax.
Therefore,
y² = 4(3)x
y² = 12x is the required equation of parabola.

10.

Since, vertex is (0, 0) and focus is (–2, 0) which lies on x-axis of parabola.
Therefore, the equation of parabola will be in the form of y² = – 4ax.
Therefore,
y² = – 4(2)x
y² = – 8x is the required equation of parabola.

11.

Since, parabola is along x-axis. So, its possible equation may be either y² = 4ax or y² = – 4ax.
Since, parabola is passing through the point (2, 3), which lies in first quadrant.
So, the equation is of the form y² = 4ax.
The point (2, 3) lies on y² = 4ax, then
(3)² = 4a(2)
9 = 8a
a = 9/8
Therefore, the equation of the parabola is
y² = 4(9/8)x
= (9/2)x
Or 2y² = 9x, which is required equation.

12.

Since, parabola is along y-axis. So, its possible equation may be either x² = 4ay or x² = – 4ay.
Since, parabola is passing through the point (5, 2),
which lies in first quadrant. So, the equation is of the form x² = 4ay.
The point (5, 2) lies on x² = 4ay, then
(5)² = 4a(2)
25 = 8a
a = 25/8
Therefore, the equation of the parabola is
x² = 4(25/8)y
= (25/2)y
Or 2x² = 25y, which is required equation.

Q.11 In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

1.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{36}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{16}}\mathbf{=}\mathbf{1}$

2.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{25}}\mathbf{=}\mathbf{1}$

3.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{16}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{9}}\mathbf{=}\mathbf{1}$

4.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{25}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{100}}\mathbf{=}\mathbf{1}$

5.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{49}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{36}}\mathbf{=}\mathbf{1}$

6.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{100}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{400}}\mathbf{=}\mathbf{1}$

7. 36x² + 4y² = 144
8. 16x² + y² = 16
9. 4x² + 9y² = 36

Ans

$\begin{array}{l}\text{Since},\text{the denominator of}\frac{{\text{x}}^{\text{2}}}{36}\text{\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{y}}^{\text{2}}}{16}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\text{we get}\\ {\mathrm{a}}^2=36{\text{and b}}^{\text{2}}=16\\ \Rightarrow \mathrm{a}=\pm 6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=\pm \text{4}\\ \text{\hspace{0.17em}c}=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\\ =\sqrt{36-16}\\ \text{\hspace{0.17em}c}=\pm \sqrt{20}\\ \mathrm{Coordinates}\text{of foci}=(\pm \sqrt{20},0)\\ \mathrm{Coordinates}\text{of vertices}=(\pm 6,0)\\ \mathrm{Length}\text{of major axis}=2\mathrm{a}\\ =2\times 6\\ =12\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ =2\times 4\\ =8\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\end{array}$ $\begin{array}{l}=\frac{\sqrt{20}}{6}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(4\right)}^2}{6}\\ =\frac{16}{3}\end{array}$

$\begin{array}{l}\mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ =2\times 2\\ =4\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{\sqrt{21}}{5}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(2\right)}^2}{5}\\ =\frac{8}{5}\end{array}$

3.

$\begin{array}{l}\text{Since},\text{the denominator of}\frac{{\text{x}}^{\text{2}}}{16}\text{\hspace{0.17em}\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{y}}^{\text{2}}}{9}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\text{we get}\\ {\mathrm{a}}^2=16{\text{and b}}^{\text{2}}=9\\ \Rightarrow \mathrm{a}=\pm 4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=\pm \text{\hspace{0.17em}3}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\\ =\sqrt{16-9}\\ \text{\hspace{0.17em} c}=\pm \sqrt{7}\end{array}$ $\begin{array}{l}\mathrm{Coordinates}\text{of foci}=(\pm \sqrt{7},0)\\ \mathrm{Coordinates}\text{of vertices}=(\pm 4,0)\\ \mathrm{Length}\text{of major axis}=2\mathrm{a}\\ =2\times 4\\ =8\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ =2\times 3\\ =6\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{\sqrt{7}}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(3\right)}^2}{4}\\ =\frac{9}{2}\end{array}$

$\begin{array}{l}\text{Since},\text{the denominator of}\frac{{\text{y}}^{\text{2}}}{100}\text{\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{x}}^{\text{2}}}{25}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1,\text{we get}\\ {\mathrm{a}}^2=100{\text{and b}}^{\text{2}}=25\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{a}=\pm 10\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=\pm \text{\hspace{0.17em}5}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\\ =\sqrt{100-25}\\ \text{\hspace{0.17em} c}=\pm \sqrt{75}=\pm 5\sqrt{3}\\ \mathrm{Coordinates}\text{of foci}=(0,\pm 5\sqrt{3})\\ \mathrm{Coordinates}\text{of vertices}=(0,\pm 10)\\ \mathrm{Length}\text{of major axis}=2\mathrm{a}\\ =2\times 10\\ =20\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ =2\times 5\\ =10\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{5\sqrt{3}}{10}\\ =\frac{\sqrt{3}}{2}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(5\right)}^2}{10}\\ =5\end{array}$

$\text{Since},\text{the denominator of}\frac{{\text{x}}^{\text{2}}}{49}\text{\hspace{0.17em}\hspace{0.17em}is greater than the denominator}$ $\begin{array}{l}\text{of}\frac{{\text{y}}^{\text{2}}}{36}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\text{we get}\\ {\mathrm{a}}^2=49{\text{and b}}^{\text{2}}=36\\ \Rightarrow \mathrm{a}=\pm 7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=\pm \text{\hspace{0.17em}6}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\\ =\sqrt{49-36}\\ \text{\hspace{0.17em} c}=\pm \sqrt{13}\\ \mathrm{Coordinates}\text{of foci}=(\pm \sqrt{13},0)\\ \mathrm{Coordinates}\text{of vertices}=(\pm 7,0)\\ \mathrm{Length}\text{of major axis}=2\mathrm{a}\\ =2\times 7\\ =14\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ =2\times 6\\ =12\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{\sqrt{13}}{7}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(6\right)}^2}{7}\end{array}$ $\mathrm{Length}\text{of latus rectum}=\frac{72}{7}$

$\begin{array}{l}\text{Since},\text{the denominator of}\frac{{\text{y}}^{\text{2}}}{400}\text{\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{x}}^{\text{2}}}{100}.\text{Comparing the given equation with}\\ \frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1,\text{we get}\\ {\mathrm{a}}^2=400{\text{and b}}^{\text{2}}=100\\ \Rightarrow \mathrm{a}=\pm 20\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=\pm \text{\hspace{0.17em}10}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\\ =\sqrt{400-100}\\ \text{\hspace{0.17em} c}=\pm \sqrt{300}=\pm 10\sqrt{3}\\ \mathrm{Coordinates}\text{of foci}=(0,\pm 10\sqrt{3})\\ \mathrm{Coordinates}\text{of vertices}=(0,\pm 20)\\ \mathrm{Length}\text{of major axis}=2\mathrm{a}\\ =2\times 20\\ =40\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ =2\times 10\\ =20\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\end{array}$ $\begin{array}{l}=\frac{10\sqrt{3}}{20}\\ =\frac{\sqrt{3}}{2}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(10\right)}^2}{20}\\ =10\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of ellipse is:}\\ 36{\mathrm{x}}^2+4{\mathrm{y}}^2=144\\ \mathrm{Dividing}\text{both sides by 144, we get}\\ \frac{36{\mathrm{x}}^2}{144}+\frac{4{\mathrm{y}}^2}{144}=\frac{144}{144}\\ \Rightarrow \frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{36}=1\\ \text{Since},\text{the denominator of}\frac{{\text{y}}^{\text{2}}}{36}\text{\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{x}}^{\text{2}}}{4}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1,\text{we get}\\ {\mathrm{a}}^2=36{\text{\hspace{0.17em}and \hspace{0.17em}b}}^{\text{2}}=4\\ \Rightarrow \mathrm{a}=\pm 6\text{\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=\pm \text{\hspace{0.17em}2}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\\ =\sqrt{36-4}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}c}=\pm \sqrt{32}=\pm 4\sqrt{2}\\ \mathrm{Coordinates}\text{of foci}=(0,\pm 4\sqrt{2})\\ \mathrm{Coordinates}\text{of vertices}=(0,\pm 6)\\ \mathrm{Length}\text{of major axis}=2\mathrm{a}\\ =2\times 6\\ =12\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ =2\times 2\\ =4\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{4\sqrt{2}}{6}\\ =\frac{2\sqrt{2}}{3}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(2\right)}^2}{6}\\ =\frac{4}{3}\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of ellipse is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}16x}}^{\text{2}}+{\text{y}}^{\text{2}}=\text{16}\\ \mathrm{Dividing}\text{both sides by 144, we get}\end{array}$ $\begin{array}{l}\frac{16{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{16}=\frac{16}{16}\\ \Rightarrow \frac{{\mathrm{x}}^2}{1}+\frac{{\mathrm{y}}^2}{16}=1\\ \text{Since},\text{the denominator of}\frac{{\text{y}}^{\text{2}}}{16}\text{\hspace{0.17em}\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{x}}^{\text{2}}}{1}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1,\text{we get}\\ {\mathrm{a}}^2=16{\text{\hspace{0.17em}and \hspace{0.17em}b}}^{\text{2}}=1\\ \Rightarrow \mathrm{a}=\pm 4\text{\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=\pm \text{\hspace{0.17em}1}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\\ =\sqrt{16-1}\\ \text{\hspace{0.17em} c}=\pm \sqrt{15}\\ \mathrm{Coordinates}\text{of foci}=(0,\pm \sqrt{15})\\ \mathrm{Coordinates}\text{of vertices}=(0,\pm 4)\\ \mathrm{Length}\text{of major axis}=2\mathrm{a}\\ =2\times 4\\ =8\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ =2\times 1\\ =2\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{\sqrt{15}}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}=\frac{2{\left(1\right)}^2}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{1}{2}\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of ellipse is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 4x}}^{\text{2}}+{\text{9y}}^{\text{2}}=\text{36}\\ \mathrm{Dividing}\text{both sides by 36, we get}\\ \frac{4{\mathrm{x}}^2}{\text{36}}+\frac{9{\mathrm{y}}^2}{\text{36}}=\frac{\text{36}}{\text{36}}\\ \Rightarrow \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{4}=1\\ \text{Since},\text{the denominator of}\frac{{\text{x}}^{\text{2}}}{9}\text{\hspace{0.17em}\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{y}}^{\text{2}}}{4}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\text{we get}\\ {\mathrm{a}}^2=9{\text{\hspace{0.17em}\hspace{0.17em}and b}}^{\text{2}}=4\\ \Rightarrow \mathrm{a}=\pm 3\text{and \hspace{0.17em}\hspace{0.17em}b}=\pm \text{\hspace{0.17em}2}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\\ =\sqrt{9-4}\\ \text{\hspace{0.17em} c}=\pm \sqrt{5}\\ \mathrm{Coordinates}\text{of foci}=(\pm \sqrt{5},0)\\ \mathrm{Coordinates}\text{of vertices}=(\pm 3,0)\end{array}$ $\begin{array}{l}\mathrm{Length}\text{of major axis}=2\mathrm{a}\\ =2\times 3\\ =6\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ =2\times 2\\ =4\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{\sqrt{5}}{3}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(2\right)}^2}{3}\\ =\frac{8}{3}\end{array}$

Q.12 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Vertices}\mathbf{(}\mathbf{\pm }\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{)}$

Ans

$\begin{array}{l}\mathrm{Since},\text{vertices are on x-axis, then equation of ellipse is:}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\text{where a is semi-major axis.}\\ \text{Given: \hspace{0.17em}\hspace{0.17em}a}=5\text{and c}=\pm 4\\ \mathrm{Since},{\text{\hspace{0.17em}c}}^{\text{2}}={\mathrm{a}}^2-{\mathrm{b}}^2\\ \therefore {\text{4}}^{\text{2}}=5^2-{\mathrm{b}}^2\end{array}$ $\begin{array}{l}\Rightarrow {\mathrm{b}}^2=25-16=9\\ \mathrm{Therefore},\text{equation of ellipse is}\\ \frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{9}=1\end{array}$

Q.13 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Vertices}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{13}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{5}\mathbf{)}$

Ans

$\begin{array}{l}\mathrm{Since},\text{vertices are on y-axis, then equation of ellipse is:}\\ \frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1,\text{where a is semi-major axis.}\\ \text{Given: a}=13\text{and c}=\pm 5\\ \mathrm{Since},{\text{\hspace{0.17em}c}}^{\text{2}}={\mathrm{a}}^2-{\mathrm{b}}^2\\ \therefore {\text{5}}^{\text{2}}={13}^2-{\mathrm{b}}^2\\ \Rightarrow {\mathrm{b}}^2=169-25=144\\ \mathrm{Therefore},\text{equation of ellipse is}\\ \frac{{\mathrm{x}}^2}{144}+\frac{{\mathrm{y}}^2}{169}=1\end{array}$

Q.14 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Vertices}\mathbf{(}\mathbf{\pm }\mathbf{6}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{foci}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{)}$

Ans

$\begin{array}{l}\mathrm{Since},\text{vertices are on x-axis, then equation of ellipse is:}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\text{where a is semi-major axis.}\\ \text{Given: a}=6\text{and c}=\pm 4\\ \mathrm{Since},{\text{\hspace{0.17em}c}}^{\text{2}}={\mathrm{a}}^2-{\mathrm{b}}^2\\ \therefore {\text{4}}^{\text{2}}=6^2-{\mathrm{b}}^2\\ \Rightarrow {\mathrm{b}}^2=36-16=20\end{array}$ $\begin{array}{l}\mathrm{Therefore},\text{equation of ellipse is}\\ \frac{{\mathrm{x}}^2}{36}+\frac{{\mathrm{y}}^2}{20}=1\end{array}$

Q.15 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Ends}\mathbf{}\mathbf{of}\mathbf{}\mathbf{major}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{Ends}\mathbf{}\mathbf{of}\mathbf{}\mathbf{minor}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{2}\mathbf{)}$

Ans

$\begin{array}{l}\mathrm{Ends}\mathrm{of}\mathrm{major}\mathrm{axis}(\pm 3,0),\mathrm{Ends}\mathrm{of}\mathrm{minor}\mathrm{axis}(0,\pm 2)\\ \mathrm{Since},\text{vertices are on x-axis.Then equation of ellipse is}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\text{where a is semi-major axis.}\\ \text{Given: a}=\text{3 and b}=2\\ \mathrm{Then},\text{equation of required ellipse is:}\\ \frac{{\mathrm{x}}^2}{3^2}+\frac{{\mathrm{y}}^2}{2^2}=1\\ \Rightarrow \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{4}=1\end{array}$

Q.16 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Ends}\mathbf{}\mathbf{of}\mathbf{}\mathbf{major}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\sqrt{\mathbf{5}}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{Ends}\mathbf{}\mathbf{of}\mathbf{}\mathbf{minor}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}$

Ans

$\begin{array}{l}\mathrm{Ends}\mathrm{of}\mathrm{major}\mathrm{axis}(0,\pm \sqrt{5}),\mathrm{Ends}\mathrm{of}\mathrm{minor}\mathrm{axis}(\pm 1,0)\\ \mathrm{Since},\text{vertices are on y-axis.Then equation of ellipse is}\\ \frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1,\text{where a is semi-major axis.}\\ \text{Given: a}=\sqrt{5}\text{and b}=1\\ \mathrm{Then},\text{equation of required ellipse is:}\end{array}$ $\begin{array}{l}\frac{{\mathrm{x}}^2}{1^2}+\frac{{\mathrm{y}}^2}{{\left(\sqrt{5}\right)}^2}=1\\ \Rightarrow \frac{{\mathrm{x}}^2}{1}+\frac{{\mathrm{y}}^2}{5}=1\end{array}$

Q.17 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Length}\mathbf{}\mathbf{of}\mathbf{}\mathbf{major}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{26}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{A}$

Ans

$\begin{array}{l}\mathrm{Since},\text{foci}(\pm 5,0)\text{lies on x-axis.So, equation of ellipse is}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\text{where a is semi-major axis.}\\ \text{Given: Length of major axis}\left(2\mathrm{a}\right)=26\\ \therefore \mathrm{a}=\frac{26}{2}=13\text{and c}=5\\ âˆ\mu {\mathrm{c}}^2={\mathrm{a}}^2-{\mathrm{b}}^2\\ \Rightarrow 5^2={\left(13\right)}^2-{\mathrm{b}}^2\\ \Rightarrow 25=169-{\mathrm{b}}^2\\ \Rightarrow {\mathrm{b}}^2=169-25\\ =144\\ \mathrm{The}\text{equation of ellipse is:}\\ \frac{{\mathrm{x}}^2}{169}+\frac{{\mathrm{y}}^2}{144}=1\end{array}$

Q.18 Find the equation for the ellipse that satisfies the given condition:A

$\mathbf{Length}\mathbf{}\mathbf{of}\mathbf{}\mathbf{minor}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{16}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{6}\mathbf{)}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Length}\text{of minor axis}=16,\\ \mathrm{foci}=(0,\pm 6)\\ \mathrm{Length}\text{of semi-minor axis}=\frac{16}{2}\end{array}$ $\begin{array}{l}\mathrm{b}=8\\ \mathrm{Since},\text{foci lies on y-axis, so equation of ellipse is}\\ \frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1\\ âˆ\mu {\mathrm{c}}^2={\mathrm{a}}^2-{\mathrm{b}}^2\\ \Rightarrow 6^2={\mathrm{a}}^2-8^2\\ \Rightarrow 36={\mathrm{a}}^2-64\\ \Rightarrow {\mathrm{a}}^2=36+64\\ =100\\ \frac{{\mathrm{x}}^2}{64}+\frac{{\mathrm{y}}^2}{100}=1,\text{which is required problem.}\end{array}$

Q.19 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Foci}\mathbf{(}\mathbf{\pm }\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{a}\mathbf{=}\mathbf{4}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Length}\text{of semi-major axis, a}=4,\\ \mathrm{foci}=(\pm 3,0)\\ âˆ\mu {\mathrm{c}}^2={\mathrm{a}}^2-{\mathrm{b}}^2\\ \therefore 3^2=4^2-{\mathrm{b}}^2\\ \Rightarrow {\mathrm{b}}^2=16-9\\ =7\\ \mathrm{Since},\text{foci is on x-axis. So, equation of ellipse is}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\\ \text{Putting value of a and b in equation of ellipse, we get}\\ \frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{7}=1\end{array}$

Q.20 Find the equation for the ellipse that satisfies the given condition:
b = 3, c = 4, centre at the origin; foci on the x axis.

Ans

$\begin{array}{l}\mathrm{Since},\text{foci is on x-axis. So, equation of ellipse is}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\\ \mathrm{Given}:\mathrm{b}=3\text{and c}=4\\ \mathrm{So},{\mathrm{c}}^2={\mathrm{a}}^2-{\mathrm{b}}^2\\ \Rightarrow 4^2={\mathrm{a}}^2-3^2\\ \Rightarrow {\mathrm{a}}^2=16+9\\ =25\\ \Rightarrow \mathrm{a}=\pm 5\\ \mathrm{Therefore},\text{the equation of ellipse is:}\\ \frac{{\mathrm{x}}^2}{5^2}+\frac{{\mathrm{y}}^2}{3^2}=1\\ \Rightarrow \frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{9}=1\end{array}$

Q.21 Find the equation for the ellipse that satisfies the given condition:
Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Ans

$\begin{array}{l}\mathrm{Equation}\text{of ellipse when major axis lies on y-axis.}\\ \frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1...\left(\mathrm{i}\right)\\ \mathrm{where}\text{a is semi-major axis.}\\ \mathrm{Since},\text{ellipse}\left(\mathrm{i}\right)\text{passes through the point}(3,2).\mathrm{So},\\ \frac{3^2}{{\mathrm{b}}^2}+\frac{2^2}{{\mathrm{a}}^2}=1\end{array}$ $\begin{array}{l}\Rightarrow \frac{9}{{\mathrm{b}}^2}+\frac{4}{{\mathrm{a}}^2}=1...\left(\mathrm{ii}\right)\\ \mathrm{Since},\text{ellipse}\left(\mathrm{i}\right)\text{passes through the point}(1,6).\mathrm{So},\\ \frac{1^2}{{\mathrm{b}}^2}+\frac{6^2}{{\mathrm{a}}^2}=1\\ \Rightarrow \frac{1}{{\mathrm{b}}^2}+\frac{36}{{\mathrm{a}}^2}=1...\left(\mathrm{iii}\right)\\ \mathrm{Subtracting}\text{equation}\left(\mathrm{ii}\right)\text{from 9}\times \text{equation}\left(\mathrm{iii}\right),\mathrm{we}\text{\hspace{0.17em}get}\\ \frac{9}{{\mathrm{b}}^2}+\frac{324}{{\mathrm{a}}^2}=9\\ \underset{¯}{\pm \frac{9}{{\mathrm{b}}^2}\pm \frac{4}{{\mathrm{a}}^2}=\pm 1}\\ \frac{320}{{\mathrm{a}}^2}=8\\ \Rightarrow {\mathrm{a}}^2=\frac{320}{8}=40\\ \mathrm{Putting}{\text{value of a}}^{\text{2}}\text{in equation}\left(\mathrm{iii}\right),\text{we get}\\ \frac{1}{{\mathrm{b}}^2}+\frac{36}{40}=1\\ \Rightarrow \frac{1}{{\mathrm{b}}^2}=1-\frac{36}{40}\\ =\frac{4}{40}\\ \Rightarrow {\mathrm{b}}^2=10\\ \mathrm{Putting}{\text{value of a}}^{\text{2}}{\text{and b}}^{\text{2}}\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \frac{{\mathrm{x}}^2}{10}+\frac{{\mathrm{y}}^2}{40}=1\end{array}$

Q.22 Find the equation for the ellipse that satisfies the given condition:
Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

Ans

$\begin{array}{l}\mathrm{Equation}\text{of ellipse when major axis lies on y-axis.}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1...\left(\mathrm{i}\right)\\ \mathrm{where}\text{a is semi-major axis.}\\ \mathrm{Since},\text{ellipse}\left(\mathrm{i}\right)\text{passes through the point}(4,3).\mathrm{So},\\ \frac{4^2}{{\mathrm{a}}^2}+\frac{3^2}{{\mathrm{b}}^2}=1\\ \Rightarrow \frac{16}{{\mathrm{a}}^2}+\frac{9}{{\mathrm{b}}^2}=1...\left(\mathrm{ii}\right)\\ \mathrm{Since},\text{ellipse}\left(\mathrm{i}\right)\text{passes through the point}(6,2).\mathrm{So},\\ \frac{6^2}{{\mathrm{a}}^2}+\frac{2^2}{{\mathrm{b}}^2}=1\\ \Rightarrow \frac{36}{{\mathrm{a}}^2}+\frac{4}{{\mathrm{b}}^2}=1...\left(\mathrm{iii}\right)\\ \mathrm{Subtracting}\text{4}\times \text{equation}\left(\mathrm{ii}\right)\text{from 9}\times \text{equation}\left(\mathrm{iii}\right),\mathrm{we}\text{\hspace{0.17em}get}\\ \frac{324}{{\mathrm{a}}^2}+\frac{36}{{\mathrm{b}}^2}=9\\ \underset{¯}{\pm \frac{64}{{\mathrm{a}}^2}\pm \frac{36}{{\mathrm{b}}^2}=\pm 4}\\ \frac{260}{{\mathrm{a}}^2}=5\\ \Rightarrow {\mathrm{a}}^2=\frac{260}{5}=52\end{array}$ $\begin{array}{l}\mathrm{Putting}{\text{value of a}}^{\text{2}}\text{in equation}\left(\mathrm{iii}\right),\text{we get}\\ \frac{36}{52}+\frac{4}{{\mathrm{b}}^2}=1\\ \Rightarrow \frac{4}{{\mathrm{b}}^2}=1-\frac{36}{52}\\ =\frac{16}{52}\\ =\frac{4}{13}\\ \Rightarrow {\mathrm{b}}^2=13\\ \mathrm{Putting}{\text{value of a}}^{\text{2}}{\text{and b}}^{\text{2}}\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \frac{{\mathrm{x}}^2}{52}+\frac{{\mathrm{y}}^2}{13}=1\end{array}$

Q.23 In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

Ans

$\begin{array}{l}\mathrm{Comparing}\mathrm{}\frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1\mathrm{with}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\mathrm{we}\mathrm{get}\\ {\mathrm{a}}^{\mathrm{2}}=16\mathrm{and}{\mathrm{b}}^{\mathrm{2}}=9\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{a}=\pm 4\mathrm{and}\mathrm{b}=\pm 3\\ \mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{c}=\pm \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}=\pm \sqrt{16+9}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}=\pm \sqrt{25}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{c}=\pm 5\\ \mathrm{Coordinates}\text{of foci are}(\pm 5,0).\\ \mathrm{Coordinates}\text{of vertices are}(\pm 4,0).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{5}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(3\right)}^2}{4}\\ =\frac{9}{2}\end{array}$

$\begin{array}{l}\mathrm{Comparing}\frac{{\mathrm{y}}^2}{9}-\frac{{\mathrm{x}}^2}{27}=1\text{with}\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1,\text{we get}\\ {\text{a}}^{\text{2}}=9{\text{and b}}^{\text{2}}=27\\ \Rightarrow \mathrm{a}=\pm 3\text{and b}=\pm 3\sqrt{3}\\ \mathrm{c}=\pm \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ =\pm \sqrt{9+27}\\ =\pm \sqrt{36}\\ \mathrm{c}=\pm 6\\ \mathrm{Coordinates}\text{of foci are}(0,\pm 6).\end{array}$ $\begin{array}{l}\mathrm{Coordinates}\text{of vertices are}(0,\pm 3).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{6}{3}=2\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2\times 27}{3}\\ =18\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of hyperbola is:}\\ {\text{9y}}^{\text{2}}-{\text{4x}}^{\text{2}}=\text{36}\\ \Rightarrow \frac{{\text{9y}}^{\text{2}}}{36}-\frac{{\text{4x}}^{\text{2}}}{36}=\frac{\text{36}}{36}\\ \Rightarrow \frac{{\text{y}}^{\text{2}}}{4}-\frac{{\text{x}}^{\text{2}}}{9}=1...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\frac{{\text{y}}^{\text{2}}}{4}-\frac{{\text{x}}^{\text{2}}}{9}=1\text{with}\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1,\text{we get}\\ {\text{a}}^{\text{2}}=4{\text{and b}}^{\text{2}}=9\\ \Rightarrow \mathrm{a}=\pm 2\text{and b}=\pm 3\\ \mathrm{c}=\pm \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ =\pm \sqrt{4+9}\\ \mathrm{c}=\pm \sqrt{13}\\ \mathrm{Coordinates}\text{of foci are}(0,\pm \sqrt{13}).\end{array}$ $\begin{array}{l}\mathrm{Coordinates}\text{of vertices are}(0,\pm 2).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{\sqrt{13}}{2}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2\times 9}{2}\\ =9\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of hyperbola is:}\\ 16{\mathrm{x}}^2-9{\mathrm{y}}^2=576\\ \Rightarrow \frac{16{\mathrm{x}}^2}{576}-\frac{9{\mathrm{y}}^2}{576}=\frac{576}{576}\\ \Rightarrow \frac{{\text{x}}^{\text{2}}}{36}-\frac{{\text{y}}^{\text{2}}}{64}=1...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\frac{{\text{x}}^{\text{2}}}{36}-\frac{{\text{y}}^{\text{2}}}{64}=1\text{with}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\text{we get}\\ {\text{a}}^{\text{2}}=36{\text{and b}}^{\text{2}}=64\\ \Rightarrow \mathrm{a}=\pm 6\text{and b}=\pm 8\\ \mathrm{c}=\pm \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ =\pm \sqrt{36+64}\\ \mathrm{c}=\pm \sqrt{100}\\ =\pm 10\\ \mathrm{Coordinates}\text{of foci are}(\pm 10,0).\end{array}$ $\begin{array}{l}\mathrm{Coordinates}\text{of vertices are}(\pm 6,0).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{10}{6}\\ =\frac{5}{3}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(8\right)}^2}{6}\\ =\frac{64}{3}\end{array}$

$\begin{array}{l}\mathrm{The}\text{given equation of hyperbola is:}\\ 5{\mathrm{y}}^2-9{\mathrm{x}}^2=36\\ \Rightarrow \frac{{\text{5y}}^{\text{2}}}{36}-\frac{{\text{9x}}^{\text{2}}}{36}=\frac{\text{36}}{36}\\ \Rightarrow \frac{{\text{y}}^{\text{2}}}{\left(\frac{36}{5}\right)}-\frac{{\text{x}}^{\text{2}}}{4}=1...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\frac{{\text{y}}^{\text{2}}}{\left(\frac{36}{5}\right)}-\frac{{\text{x}}^{\text{2}}}{4}=1\text{with}\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1,\text{we get}\\ {\text{a}}^{\text{2}}=\frac{36}{5}{\text{and b}}^{\text{2}}=4\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{a}=\pm \frac{6}{\sqrt{5}}\text{and b}=\pm 2\\ \mathrm{c}=\pm \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ =\pm \sqrt{\frac{36}{5}+4}\\ \mathrm{c}=\pm \sqrt{\frac{36+20}{5}}\\ =\pm \sqrt{\frac{56}{5}}=\pm \frac{2\sqrt{14}}{\sqrt{5}}\\ \mathrm{Coordinates}\text{of foci are}(0,\pm \frac{2\sqrt{14}}{\sqrt{5}}).\\ \mathrm{Coordinates}\text{of vertices are}(0,\pm \frac{6}{\sqrt{5}}).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{\left(\frac{2\sqrt{14}}{\sqrt{5}}\right)}{\left(\frac{6}{\sqrt{5}}\right)}\\ =\frac{\sqrt{14}}{3}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2\times 2^2}{\left(\frac{6}{\sqrt{5}}\right)}\end{array}$ $$ $\mathrm{Length}\text{of latus rectum}=\frac{4\sqrt{5}}{3}$

$\begin{array}{l}\mathrm{The}\text{given equation of hyperbola is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 49y}}^{\text{2}}-{\text{16x}}^{\text{2}}=\text{784}\\ \Rightarrow \frac{{\text{49y}}^{\text{2}}}{\text{784}}-\frac{{\text{16x}}^{\text{2}}}{\text{784}}=\frac{\text{784}}{\text{784}}\\ \Rightarrow \frac{{\text{y}}^{\text{2}}}{16}-\frac{{\text{x}}^{\text{2}}}{49}=1...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\frac{{\text{y}}^{\text{2}}}{16}-\frac{{\text{x}}^{\text{2}}}{49}=1\text{with}\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1,\text{we get}\\ {\text{a}}^{\text{2}}=16{\text{and b}}^{\text{2}}=49\\ \Rightarrow \mathrm{a}=\pm 4\text{and b}=\pm 7\\ \mathrm{c}=\pm \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ =\pm \sqrt{16+49}\\ \mathrm{c}=\pm \sqrt{65}\\ \mathrm{Coordinates}\text{of foci are}(0,\pm \sqrt{65}).\\ \mathrm{Coordinates}\text{of vertices are}(0,\pm 4).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{\sqrt{65}}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\end{array}$ $\begin{array}{l}=\frac{2\times 7^2}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{49}{2}\end{array}$

Q.24 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Vertices}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}$

Ans

$\begin{array}{l}\mathrm{Since},\text{foci is on x-axis, the equation of hyperbola is of the form}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\\ \mathrm{Since},\mathrm{vertices}(\pm 2,0),\mathrm{foci}(\pm 3,0)\\ \mathrm{So},\text{a}=\text{2 and c}=3.\\ âˆ\mu {\mathrm{b}}^2={\mathrm{c}}^2-{\mathrm{a}}^2\\ \Rightarrow {\mathrm{b}}^2=3^2-2^2\\ =9-4\\ =5\\ \Rightarrow \mathrm{b}=\sqrt{5}\\ \mathrm{Therefore},\text{equation of hyperbola is}\\ \frac{{\mathrm{x}}^2}{2^2}-\frac{{\mathrm{y}}^2}{{\left(\sqrt{5}\right)}^2}=1\\ \Rightarrow \frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{5}=1\end{array}$

Q.25 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Vertices}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{5}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{8}\mathbf{)}$

Ans

$\mathrm{Since},\text{foci is on y-axis, the equation of hyperbola is of the form}$ $\begin{array}{l}\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1\\ \mathrm{Since},\mathrm{vertices}(0,\pm 5),\mathrm{foci}(0,\pm 8)\\ \mathrm{So},\text{a}=\text{5 and c}=8.\\ âˆ\mu {\mathrm{b}}^2={\mathrm{c}}^2-{\mathrm{a}}^2\\ \Rightarrow {\mathrm{b}}^2=8^2-5^2\\ =64-25\\ =39\\ \Rightarrow \mathrm{b}=\sqrt{39}\\ \mathrm{Therefore},\text{equation of hyperbola is}\\ \frac{{\mathrm{y}}^2}{5^2}-\frac{{\mathrm{x}}^2}{{\left(\sqrt{39}\right)}^2}=1\\ \Rightarrow \frac{{\mathrm{y}}^2}{25}-\frac{{\mathrm{x}}^2}{39}=1\end{array}$

Q.26 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Vertices}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{3}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{5}\mathbf{)}$

Ans

$\begin{array}{l}\mathrm{Since},\text{foci is on y-axis, the equation of hyperbola is of the form}\\ \frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1\\ \mathrm{Since},\mathrm{vertices}(0,\pm 3),\mathrm{foci}(0,\pm 5)\\ \mathrm{So},\text{a}=\text{3 and c}=5.\\ âˆ\mu {\mathrm{b}}^2={\mathrm{c}}^2-{\mathrm{a}}^2\\ \Rightarrow {\mathrm{b}}^2=5^2-3^2\\ =25-9\\ =16\\ \Rightarrow \mathrm{b}=\pm 4\end{array}$ $\begin{array}{l}\mathrm{Therefore},\text{equation of hyperbola is}\\ \frac{{\mathrm{y}}^2}{3^2}-\frac{{\mathrm{x}}^2}{4^2}=1\\ \Rightarrow \frac{{\mathrm{y}}^2}{9}-\frac{{\mathrm{x}}^2}{16}=1\end{array}$

Q.27 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Foci}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{the}\mathbf{}\mathbf{transverse}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{is}\mathbf{}\mathbf{of}\mathbf{}\mathbf{length}\mathbf{}\mathbf{8}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Foci}(\pm 5,0),\mathrm{the}\mathrm{transverse}\mathrm{axis}\mathrm{is}\mathrm{of}\mathrm{length}8\\ \mathrm{Since},\text{foci is on x-axis, the equation of hyperbola is of}\\ \text{the form}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\\ \mathrm{Here},\mathrm{c}=5\text{and a}=\frac{8}{2}=4\\ âˆ\mu {\mathrm{b}}^2={\mathrm{c}}^2-{\mathrm{a}}^2\\ =5^2-4^2\\ =25-16\\ =9\\ \Rightarrow \mathrm{b}=\pm 3\\ \mathrm{Therefore},\text{the equation of hyperbola is}\\ \frac{{\mathrm{x}}^2}{4^2}-\frac{{\mathrm{y}}^2}{3^2}=1\\ \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1\end{array}$

Q.28 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Foci}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{13}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{the}\mathbf{}\mathbf{conjugate}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{is}\mathbf{}\mathbf{of}\mathbf{}\mathbf{length}\mathbf{}\mathbf{24}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Foci}(0,\pm 13),\mathrm{the}\mathrm{conjugate}\mathrm{axis}\mathrm{is}\mathrm{of}\mathrm{length}24.\\ \mathrm{Since},\text{foci is on y-axis, the equation of hyperbola is of}\end{array}$ $\begin{array}{l}\text{the form}\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1\\ \mathrm{Here},\mathrm{c}=13\text{and b}=\frac{24}{2}=12\\ âˆ\mu {\mathrm{a}}^2={\mathrm{c}}^2-{\mathrm{b}}^2\\ ={13}^2-{\left(12\right)}^2\\ =169-144\\ =25\\ \Rightarrow \mathrm{a}=\pm 5\\ \mathrm{Therefore},\text{the equation of hyperbola is}\\ \frac{{\mathrm{y}}^2}{5^2}-\frac{{\mathrm{x}}^2}{{\left(12\right)}^2}=1\\ \frac{{\mathrm{y}}^2}{25}-\frac{{\mathrm{x}}^2}{144}=1\end{array}$

Q.29 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Foci}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{3}\sqrt{\mathbf{5}}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{the}\mathbf{}\mathbf{latus}\mathbf{}\mathbf{rectum}\mathbf{}\mathbf{is}\mathbf{}\mathbf{of}\mathbf{}\mathbf{length}\mathbf{}\mathbf{8}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Foci}(\pm 3\sqrt{5},0),\mathrm{the}\mathrm{latus}\mathrm{rectum}\mathrm{is}\mathrm{of}\mathrm{length}8.\\ \mathrm{Since},\text{foci is on x-axis, the equation of hyperbola is of}\\ \text{the form\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\\ \mathrm{Here},\mathrm{c}=3\sqrt{5}\text{and}\frac{{\text{2b}}^{\text{2}}}{\mathrm{a}}=8\\ \Rightarrow {\mathrm{b}}^2=4\mathrm{a}\\ âˆ\mu {\mathrm{b}}^2={\mathrm{c}}^2-{\mathrm{a}}^2\\ 4\mathrm{a}={\left(3\sqrt{5}\right)}^2-{\mathrm{a}}^2\end{array}$ $\begin{array}{l}4\mathrm{a}=45-{\mathrm{a}}^2\\ {\mathrm{a}}^2+4\mathrm{a}-45=0\\ \Rightarrow (\mathrm{a}+9)(\mathrm{a}-5)=0\\ \Rightarrow \mathrm{a}=5,-9\\ \Rightarrow \mathrm{a}=5(\mathrm{Neglecting}-\text{9})\\ \mathrm{and}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}^{\text{2}}=4\times 5\\ =20\\ \mathrm{Therefore},\text{the equation of hyperbola is}\\ \frac{{\mathrm{x}}^2}{5^2}-\frac{{\mathrm{y}}^2}{20}=1\\ \frac{{\mathrm{x}}^2}{25}-\frac{{\mathrm{y}}^2}{20}=1\end{array}$

Q.30 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Foci}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{the}\mathbf{}\mathbf{latus}\mathbf{}\mathbf{rectum}\mathbf{}\mathbf{is}\mathbf{}\mathbf{of}\mathbf{}\mathbf{length}\mathbf{}\mathbf{12}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Foci}(\pm 4,0),\mathrm{the}\mathrm{latus}\mathrm{rectum}\mathrm{is}\mathrm{of}\mathrm{length}12.\\ \mathrm{Since},\text{foci is on x-axis, the equation of hyperbola is of}\\ \text{the form\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\\ \mathrm{Here},\mathrm{c}=4\text{and}\frac{{\text{2b}}^{\text{2}}}{\mathrm{a}}=12\\ \Rightarrow {\mathrm{b}}^2=6\mathrm{a}\\ âˆ\mu {\mathrm{b}}^2={\mathrm{c}}^2-{\mathrm{a}}^2\\ 6\mathrm{a}=4^2-{\mathrm{a}}^2\\ 6\mathrm{a}=16-{\mathrm{a}}^2\\ {\mathrm{a}}^2+6\mathrm{a}-16=0\\ \Rightarrow (\mathrm{a}+8)(\mathrm{a}-2)=0\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{a}=2,-8\\ \Rightarrow \mathrm{a}=2(\mathrm{Neglecting}-\text{8})\\ \mathrm{and}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}^{\text{2}}=6\times 2[âˆ\mu {\mathrm{b}}^2=6\mathrm{a}]\\ =12\\ \mathrm{Therefore},\text{the equation of hyperbola is}\\ \frac{{\mathrm{x}}^2}{2^2}-\frac{{\mathrm{y}}^2}{12}=1\\ \frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{12}=1\end{array}$

Q.31 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Vertices}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{7}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{e}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Vertices}=(\pm 7,0),\mathrm{e}=\frac{4}{3}\\ \mathrm{Since},\text{vertices lie on x-axis, the equation of hyperbola is}\\ \text{of the form}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\\ \mathrm{Here},\text{a}=\text{7 and e}=\frac{4}{3}\\ âˆ\mu \mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}}\\ \Rightarrow \mathrm{c}=\mathrm{ae}=7\times \frac{4}{3}=\frac{28}{3}\\ \mathrm{and}{\text{\hspace{0.17em} b}}^{\text{2}}={\mathrm{c}}^2-{\mathrm{a}}^2\\ {\text{\hspace{0.17em} b}}^{\text{2}}={\left(\frac{28}{3}\right)}^2-7^2\end{array}$ $\begin{array}{l}=\frac{784}{9}-49\\ =\frac{784-441}{9}\\ =\frac{343}{9}\\ \mathrm{Therefore},\text{the equation of hyperbola is}\\ \frac{{\mathrm{x}}^2}{7^2}-\frac{{\mathrm{y}}^2}{\frac{343}{9}}=1\\ \Rightarrow \frac{{\mathrm{x}}^2}{49}-\frac{9{\mathrm{y}}^2}{343}=1\end{array}$

Q.32 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Foci}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\sqrt{\mathbf{10}}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{passing}\mathbf{}\mathbf{through}\mathbf{}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Foci}(0,\pm \sqrt{10}),\mathrm{passing}\mathrm{through}(2,3).\\ \mathrm{Since},\text{foci is on y-axis, the equation of hyperbola is of}\\ \text{the form}\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1\\ \mathrm{Here},\mathrm{c}=\sqrt{10}\\ âˆ\mu {\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2\\ \Rightarrow 10={\mathrm{a}}^2+{\mathrm{b}}^2...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{hyperbola passes through the point}(2,3).\\ \mathrm{So},\frac{3^2}{{\mathrm{a}}^2}-\frac{2^2}{{\mathrm{b}}^2}=1\\ \frac{9}{{\mathrm{a}}^2}-\frac{4}{{\mathrm{b}}^2}=1\end{array}$ $\begin{array}{l}\mathrm{Putting}{\text{b}}^{\text{2}}=10-{\mathrm{a}}^2,\text{from equation}\left(\mathrm{i}\right),\text{we get}\\ \frac{9}{{\mathrm{a}}^2}-\frac{4}{10-{\mathrm{a}}^2}=1\\ \Rightarrow \frac{9(10-{\mathrm{a}}^2)-4{\mathrm{a}}^2}{{\mathrm{a}}^2(10-{\mathrm{a}}^2)}=1\\ \Rightarrow 90-9{\mathrm{a}}^2-4{\mathrm{a}}^2=10{\mathrm{a}}^2-{\mathrm{a}}^4\\ \Rightarrow {\mathrm{a}}^4-23{\mathrm{a}}^2+90=0\\ \Rightarrow ({\mathrm{a}}^2-18)({\mathrm{a}}^2-5)=0\\ \Rightarrow {\mathrm{a}}^2=18,5\\ \mathrm{So},{\mathrm{b}}^2=10-{\mathrm{a}}^2\\ =10-5\\ =5\\ \mathrm{Neglecting}{\text{a}}^{\text{2}}=\text{18 because}{\mathrm{b}}^2=10-{\mathrm{a}}^2.\\ \mathrm{Therefore},\text{the equation of required hyperbola is}\\ \frac{{\mathrm{y}}^2}{5}-\frac{{\mathrm{x}}^2}{5}=1\end{array}$

Q.33 If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Ans

Let point A(10, 5) lies on the parabolic reflector.
Equation of parabola is x² = 4ay
Then, (10)² = 4a(5)
100 = 20a
a = 100/20
= 5

Thus, the focus of the reflector is at the mid-point of the diameter.

Q.34 An arch is in the form of a parabola with its axis vertical. The arch is 10m high and 5 m wide at the base. How wide is it 2m from the vertex of the parabola?

Ans

Let point B(2.5, – 10) lies on parabolic arch. The equation of parabola is:
x² = – 4ay …(i)
Putting x = 2.5 and y = – 10, we get
(2.5)² = – 4a (–10)
6.25 = 40 a
4a = 0.625

Then, equation of parabolic arch is
x² = – 0.625y …(ii)
Let the width of arch be 2b at height 2m from vertex. Then point A(b, – 2.5) lies on parabola,
Therefore,
b² = – 0.625(– 2)
b² = 1.25
b = 1.118
2b = 2 x 1.118
= 2.236 cm approx.
Thus, the width of arch at 2 m away from the vertex is 2.24 cm approx.

Q.35 The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Ans

Let road CF is hanging with cables. The length of the longest cable is 30 m and that of the smallest cable is 6 m. The smallest cable is 6 m long.

Road is hanging on parabolic cable.
Equation of parabola with vertex O(0, 0) is:
x² = 4ay …(i)
Here, AB = 30 – 6 = 24 m and OB = 100/2 = 50 m
The coordinates of point A lying on parabolic cable is (50, 24). Let the coordinates of point P which is 18 m away from origin be (18, h).
Point A (50, 24) lies on parabola(i), then
(50)² = 4a (24)
4a = 2500/24
= (625/6)
So, the equation of parabola is
x² = (625/6)y …(ii)
Since, point P(18, h) lies on parabola (ii), then
(18)² = (625/6)h
h = (324 x 6)/625
= 3.11 m
Therefore, length of cable which is 18 m away from the middle = 6 + 3.11
= 9.11 m.

Q.36 An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Ans

Since, width of semi-ellipse is 8 m and height of ellipse is 2 m. Coordinates of B and A are (4, 0) and (0, 2) respectively.
Here, BD = 1.5 m,
So, OD = OB – BD
= 4 – 1.5
= 2.5 m
Let height of CD be h m. Then, the coordinates of point C be (2.5, h).

$\begin{array}{l}\mathrm{Equation}\text{of ellipse is:}\\ \frac{{\mathrm{x}}^2}{4^2}+\frac{{\mathrm{y}}^2}{2^2}=1...\left(\mathrm{i}\right)\end{array}$ $\begin{array}{l}\text{Point C}(2.5,\mathrm{h})\text{lies on ellipse}\left(\mathrm{i}\right),\text{then we have}\\ \frac{{(2.5)}^2}{4^2}+\frac{{\mathrm{h}}^2}{2^2}=1\\ \Rightarrow \frac{6.25}{16}+\frac{{\mathrm{h}}^2}{4}=1\\ \Rightarrow \frac{{\mathrm{h}}^2}{4}=1-\frac{6.25}{16}\\ \Rightarrow \frac{{\mathrm{h}}^2}{4}=\frac{16-6.25}{16}\\ \Rightarrow {\mathrm{h}}^2=\frac{9.75}{16}\times 4\\ \Rightarrow \mathrm{h}=\sqrt{\frac{9.75}{16}\times 4}\\ \Rightarrow \mathrm{h}=1.56\mathrm{m}\mathrm{approx}.\\ \mathrm{Thus},\text{the height of ellipse is 1.56 m\hspace{0.17em}\hspace{0.17em}}\mathrm{approx}\text{from a point}\\ \text{which is 1.5 meter away from on end.}\end{array}$

Q.37 A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Ans

Let AB be a rod of length 12 cm. A point P divides it into two parts i.e., AP and PB of length 3 m and (12 – 3) = 9 m respectively. PQ and PR are perpendiculars on X-axis and Y-axis respectively.

$\begin{array}{l}\text{InΔPQA},\text{sin}\mathrm{\theta }=\frac{\text{y}}{3}...\left(\text{i}\right)\\ \text{InΔPRB},\text{cos}\mathrm{\theta }=\frac{\text{x}}{9}...\left(\text{ii}\right)\\ \text{Squaringandaddingequation}\left(\text{i}\right)\text{andequation}\left(\text{ii}\right)\text{,}\\ \text{weget}\\ {\text{sin}}^2\mathrm{\theta }+{\text{cos}}^2\mathrm{\theta }={\left(\frac{\text{y}}{3}\right)}^2+{\left(\frac{\text{x}}{9}\right)}^2\\ 1=\frac{{\text{y}}^2}{9}+\frac{{\text{x}}^2}{81}\\ \text{or}\frac{{\text{x}}^2}{81}+\frac{{\text{y}}^2}{9}=1,\\ \text{whichistherequiredequationofpointPonanellipse.}\end{array}$

Q.38 Find the area of the triangle formed by the lines joining the vertex of the parabola x² = 12y to the ends of its latus rectum.

Ans

Comparing x² = 12y with x² = 4ay, we get
4a = 12
a = 3
Coordinates of focus F = (0, 3)
Length of latus rectum = 4a
= 4(3)
= 12
Here, AF = FB = 12/2
= 6

$\begin{array}{l}\mathrm{Ar}\left(\mathrm{\Delta OAB}\right)=\frac{1}{2}|{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)|\\ =\frac{1}{2}|0(3-3)+6(3-0)-6(0-3)|\\ =\frac{1}{2}|0\left(0\right)+6\left(3\right)-6(-3)|\\ =18\mathrm{square}\text{units}\end{array}$

Q.39 A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Ans

Let A and B be the two flag posts at a distance of 8 m from each other. P(x, y) represents the position of a running man in such a way that:
PA + PB = 10 m

$\begin{array}{l}\mathrm{OA}=\mathrm{OB}\\ =\frac{8}{2}=4\\ \mathrm{So},\text{coordinates of A and B are}(4,0)\text{and}(-4,0)\text{respectively.}\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PA}=\sqrt{{(\mathrm{x}-4)}^2+{(\mathrm{y}-0)}^2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PB}=\sqrt{{(\mathrm{x}+4)}^2+{(\mathrm{y}-0)}^2}\\ \mathrm{According}\text{to given condition:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PA}+\mathrm{PB}=\text{10}\\ \Rightarrow \sqrt{{(\mathrm{x}-4)}^2+{(\mathrm{y}-0)}^2}+\sqrt{{(\mathrm{x}+4)}^2+{(\mathrm{y}-0)}^2}=10\end{array}$ $\begin{array}{l}\Rightarrow \sqrt{{\left(\text{x}-4\right)}^2+{\left(\text{y}-0\right)}^2}=10-\sqrt{{\left(\text{x}+4\right)}^2+{\left(\text{y}-0\right)}^2}\\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ {\text{x}}^2-8\text{x}+16+{\text{y}}^2=100+{\text{x}}^2+8\text{x}+16+{\text{y}}^2-20\sqrt{{\left(\text{x}+4\right)}^2+{\left(\text{y}-0\right)}^2}\\ 0=100+16\text{x}-20\sqrt{{\left(\text{x}+4\right)}^2+{\left(\text{y}-0\right)}^2}\\ 20\sqrt{{\left(\text{x}+4\right)}^2+{\left(\text{y}-0\right)}^2}=100+16\text{x}\\ \mathrm{Again}\mathrm{squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ 400\left({\text{x}}^2+8\text{x}+16+{\text{y}}^2\right)=10000+3200\text{x}+256{\text{x}}^2\\ \left(400-256\right){\text{x}}^2+400{\text{y}}^2=10000-6400\\ \Rightarrow 144{\text{x}}^2+400{\text{y}}^2=3600\\ \text{Dividing both sides by 3600, we get}\\ \Rightarrow \frac{144}{3600}{\text{x}}^2+\frac{400}{3600}{\text{y}}^2=\frac{3600}{3600}\\ \Rightarrow \frac{{\text{x}}^2}{25}+\frac{{\text{y}}^2}{9}=1\\ \mathrm{Which}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{posts}\mathrm{traced}\mathrm{by}\mathrm{the}\mathrm{man}.\end{array}$

Q.40 An equilateral triangle is inscribed in the parabola y² = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Ans

Let each side of equilateral triangle be 2k units.

$\begin{array}{l}\mathrm{In}\mathrm{\Delta ACO},\text{by Pythagoras theorem,}\\ \text{\hspace{0.17em}\hspace{0.17em}OC}=\sqrt{{\mathrm{OA}}^2-{\mathrm{AC}}^2}\\ =\sqrt{{\left(2\mathrm{k}\right)}^2-{\mathrm{k}}^2}\\ =\sqrt{4{\mathrm{k}}^2-{\mathrm{k}}^2}\\ =\sqrt{3}\mathrm{k}\\ \mathrm{So},\text{coordinate of point A}=(\sqrt{3}\mathrm{k},\mathrm{k})\\ {\text{Point A lies on parabola y}}^{\text{2}}=\text{4ax, then}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}k}}^{\text{2}}=\text{4a}\left(\sqrt{3}\mathrm{k}\right)\\ \Rightarrow \mathrm{k}=\text{4a}\sqrt{3}\\ \mathrm{So},\text{each side of triangle}\\ =2\mathrm{k}=2\left(\text{4a}\sqrt{3}\right)\\ =8\sqrt{3}\mathrm{a}\\ \mathrm{Thus},\text{each side of equilateral triangle is}8\sqrt{3}\mathrm{a}.\end{array}$