NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.3) Exercise 11.3

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.3) Exercise 11.3

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.3

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Q.1 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.

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

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

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

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

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

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

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

Ans.

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

$\begin{array}{l} = \frac{\sqrt{20}}{6} \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 {(4)}^{2}}{6} \\ = \frac{16}{3} \end{array}$

\begin{array}{l} Since, the denominator of \frac{y^{2}}{25} is greater than the denominator \\ of \frac{x^{2}}{4} . Comparing the given equation with \\ \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1, we get \\ a^{2} = 25 {and b}^{2} = 4 \\ \Rightarrow a = \pm 5 and   b = \pm 2 \\ c = \sqrt{a^{2} - b^{2}} \\ = \sqrt{25 - 4} \\ c = \pm \sqrt{21} \\ Coordinates of foci = (0, \pm \sqrt{21}) \\ Coordinates of vertices = (0, \pm 5) \\ Length of major axis = 2 a \\ = 2 \times 5 \\ = 10 \end{array}

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

$\begin{array}{l} Since, the denominator of \frac{x^{2}}{16} is greater than the denominator \\ of \frac{y^{2}}{9} . Comparing the given equation with \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, we get \\ a^{2} = 16 {and b}^{2} = 9 \\ \Rightarrow a = \pm 4 and b = \pm 3 \\ c = \sqrt{a^{2} - b^{2}} \\ = \sqrt{16 - 9} \\ c = \pm \sqrt{7} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} Coordinates of foci = (\pm \sqrt{7}, 0) \\ Coordinates of vertices = (\pm 4, 0) \\ Length of major axis = 2 a \\ = 2 \times 4 \\ = 8 \\ Length of minor axis = 2 b \\ = 2 \times 3 \\ = 6 \\ Eccentricity = \frac{c}{a} \\ = \frac{\sqrt{7}}{4} \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 {(3)}^{2}}{4} \\ = \frac{9}{2} \end{array}$

$\begin{array}{l} Since, the denominator of \frac{y^{2}}{100} is greater than the denominator \\ of \frac{x^{2}}{25} . Comparing the given equation with \\ \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1, we get \\ a^{2} = 100 {and b}^{2} = 25 \end{array}$

$\begin{array}{l} \Rightarrow a = \pm 10 and b = \pm 5 \\ c = \sqrt{a^{2} - b^{2}} \\ = \sqrt{100 - 25} \\ c = \pm \sqrt{75} = \pm 5 \sqrt{3} \\ Coordinates of foci = (0, \pm 5 \sqrt{3}) \\ Coordinates of vertices = (0, \pm 10) \\ Length of major axis = 2 a \\ = 2 \times 10 \\ = 20 \\ Length of minor axis = 2 b \\ = 2 \times 5 \\ = 10 \\ Eccentricity = \frac{c}{a} \\ = \frac{5 \sqrt{3}}{10} \\ = \frac{\sqrt{3}}{2} \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 {(5)}^{2}}{10} \\ = 5 \end{array}$

$Since, the denominator of \frac{x^{2}}{49} is greater than the denominator$

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

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

$\begin{array}{l} = \frac{10 \sqrt{3}}{20} \\ = \frac{\sqrt{3}}{2} \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 {(10)}^{2}}{20} \\ = 10 \end{array}$

$\begin{array}{l} The given equation of ellipse is: \\ 36 x^{2} + 4 y^{2} = 144 \\ Dividing both sides by 144, we get \\ \frac{36 x^{2}}{144} + \frac{4 y^{2}}{144} = \frac{144}{144} \\ \Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{36} = 1 \\ Since, the denominator of \frac{y^{2}}{36} is greater than the denominator \\ of \frac{x^{2}}{4} . Comparing the given equation with \\ \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1, we get \\ a^{2} = 36 {and b}^{2} = 4 \\ \Rightarrow a = \pm 6 and b = \pm 2 \\ c = \sqrt{a^{2} - b^{2}} \\ = \sqrt{36 - 4} \end{array}$

$\begin{array}{l} c = \pm \sqrt{32} = \pm 4 \sqrt{2} \\ Coordinates of foci = (0, \pm 4 \sqrt{2}) \\ Coordinates of vertices = (0, \pm 6) \\ Length of major axis = 2 a \\ = 2 \times 6 \\ = 12 \\ Length of minor axis = 2 b \\ = 2 \times 2 \\ = 4 \\ Eccentricity = \frac{c}{a} \\ = \frac{4 \sqrt{2}}{6} \\ = \frac{2 \sqrt{2}}{3} \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 {(2)}^{2}}{6} \\ = \frac{4}{3} \end{array}$

$\begin{array}{l} The given equation of ellipse is: \\ {16x}^{2} + y^{2} = 16 \\ Dividing both sides by 144, we get \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} \frac{16 x^{2}}{16} + \frac{y^{2}}{16} = \frac{16}{16} \\ \Rightarrow \frac{x^{2}}{1} + \frac{y^{2}}{16} = 1 \\ Since, the denominator of \frac{y^{2}}{16} is greater than the denominator \\ of \frac{x^{2}}{1} . Comparing the given equation with \\ \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1, we get \\ a^{2} = 16 {and b}^{2} = 1 \\ \Rightarrow a = \pm 4 and b = \pm 1 \\ c = \sqrt{a^{2} - b^{2}} \\ = \sqrt{16 - 1} \\ c = \pm \sqrt{15} \\ Coordinates of foci = (0, \pm \sqrt{15}) \\ Coordinates of vertices = (0, \pm 4) \\ Length of major axis = 2 a \\ = 2 \times 4 \\ = 8 \\ Length of minor axis = 2 b \\ = 2 \times 1 \\ = 2 \\ Eccentricity = \frac{c}{a} \\ = \frac{\sqrt{15}}{4} \\ Length of latus rectum = \frac{2 b^{2}}{a} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{2 {(1)}^{2}}{4} \\ Length of latus rectum = \frac{1}{2} \end{array}$

$\begin{array}{l} The given equation of ellipse is: \\ {4x}^{2} + {9y}^{2} = 36 \\ Dividing both sides by 36, we get \\ \frac{4 x^{2}}{36} + \frac{9 y^{2}}{36} = \frac{36}{36} \\ \Rightarrow \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \\ Since, the denominator of \frac{x^{2}}{9} is greater than the denominator \\ of \frac{y^{2}}{4} . Comparing the given equation with \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, we get \\ a^{2} = 9 {and b}^{2} = 4 \\ \Rightarrow a = \pm 3 and b = \pm 2 \\ c = \sqrt{a^{2} - b^{2}} \\ = \sqrt{9 - 4} \\ c = \pm \sqrt{5} \\ Coordinates of foci = (\pm \sqrt{5}, 0) \\ Coordinates of vertices = (\pm 3, 0) \end{array}$

$\begin{array}{l} Length of major axis = 2 a \\ = 2 \times 3 \\ = 6 \\ Length of minor axis = 2 b \\ = 2 \times 2 \\ = 4 \\ Eccentricity = \frac{c}{a} \\ = \frac{\sqrt{5}}{3} \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 {(2)}^{2}}{3} \\ = \frac{8}{3} \end{array}$

Q.2 Find the equation for the ellipse that satisfies the given condition:

$Vertices (\pm 5, 0), foci (\pm 4, 0)$

Ans.

$\begin{array}{l} Since, vertices are on x-axis, then equation of ellipse is: \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a is semi-major axis. \\ Given: a = 5 and c = \pm 4 \\ Since, c^{2} = a^{2} - b^{2} \\ ∴ 4^{2} = 5^{2} - b^{2} \end{array}$

$\begin{array}{l} \Rightarrow b^{2} = 25 - 16 = 9 \\ Therefore, equation of ellipse is \\ \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 \end{array}$

Q.3 Find the equation for the ellipse that satisfies the given condition:

$Vertices (0, \pm 13), foci (0, \pm 5)$

Ans.

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

Q.4 Find the equation for the ellipse that satisfies the given condition:

$Vertices (\pm 6, 0), foci (\pm 4, 0)$

Ans.

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

$\begin{array}{l} Therefore, equation of ellipse is \\ \frac{x^{2}}{36} + \frac{y^{2}}{20} = 1 \end{array}$

Q.5 Find the equation for the ellipse that satisfies the given condition:

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

Ans.

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

Q.6 Find the equation for the ellipse that satisfies the given condition:

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

Ans.

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

$\begin{array}{l} \frac{x^{2}}{1^{2}} + \frac{y^{2}}{{(\sqrt{5})}^{2}} = 1 \\ \Rightarrow \frac{x^{2}}{1} + \frac{y^{2}}{5} = 1 \end{array}$

Q.7 Find the equation for the ellipse that satisfies the given condition:

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

Ans.

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

Q.8 Find the equation for the ellipse that satisfies the given condition:A

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

Ans.

$\begin{array}{l} Given : Length of minor axis = 16, \\ foci = (0, \pm 6) \\ Length of semi-minor axis = \frac{16}{2} \end{array}$

$\begin{array}{l} b = 8 \\ Since, foci lies on y-axis, so equation of ellipse is \\ \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \\ ∵ c^{2} = a^{2} - b^{2} \\ \Rightarrow 6^{2} = a^{2} - 8^{2} \\ \Rightarrow 36 = a^{2} - 64 \\ \Rightarrow a^{2} = 36 + 64 \\ = 100 \\ \frac{x^{2}}{64} + \frac{y^{2}}{100} = 1, which is required problem. \end{array}$

Q.9 Find the equation for the ellipse that satisfies the given condition:

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

Ans.

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

Q.10 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.

$\begin{array}{l} Equation of ellipse when major axis lies on y-axis. \\ \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 . .. (i) \\ where a is semi-major axis. \\ Since, ellipse (i) passes through the point (3, 2) . So, \\ \frac{3^{2}}{b^{2}} + \frac{2^{2}}{a^{2}} = 1 \end{array}$

$\begin{array}{l} \Rightarrow \frac{9}{b^{2}} + \frac{4}{a^{2}} = 1 . .. (ii) \\ Since, ellipse (i) passes through the point (1, 6) . So, \\ \frac{1^{2}}{b^{2}} + \frac{6^{2}}{a^{2}} = 1 \\ \Rightarrow \frac{1}{b^{2}} + \frac{36}{a^{2}} = 1 . .. (iii) \\ Subtracting equation (ii) from 9 \times equation (iii), we get \\ \frac{9}{b^{2}} + \frac{324}{a^{2}} = 9 \\ \underline{\pm \frac{9}{b^{2}} \pm \frac{4}{a^{2}} = \pm 1} \\ \frac{320}{a^{2}} = 8 \\ \Rightarrow a^{2} = \frac{320}{8} = 40 \\ Putting {value of a}^{2} in equation (iii), we get \\ \frac{1}{b^{2}} + \frac{36}{40} = 1 \\ \Rightarrow \frac{1}{b^{2}} = 1 - \frac{36}{40} \\ = \frac{4}{40} \\ \Rightarrow b^{2} = 10 \\ Putting {value of a}^{2} {and b}^{2} in equation (i), we get \\ \frac{x^{2}}{10} + \frac{y^{2}}{40} = 1 \end{array}$

Q.11 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.

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

$\begin{array}{l} Putting {value of a}^{2} in equation (iii), we get \\ \frac{36}{52} + \frac{4}{b^{2}} = 1 \\ \Rightarrow \frac{4}{b^{2}} = 1 - \frac{36}{52} \\ = \frac{16}{52} \\ = \frac{4}{13} \\ \Rightarrow b^{2} = 13 \\ Putting {value of a}^{2} {and b}^{2} in equation (i), we get \\ \frac{x^{2}}{52} + \frac{y^{2}}{13} = 1 \end{array}$

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.3) Exercise 11.3

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.3) Exercise 11.3

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4. What topics and subtopics are covered in Class 11 Mathematics?

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