NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.4) Exercise 11.4

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.4) Exercise 11.4

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Access NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.4

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Q.1 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.

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

\frac{y^{2}}{9} - \frac{x^{2}}{27} = 1

3. 9y² – 4x² = 36
4. 16 x² – 9y² = 576
5. 5y² – 9x² = 36
6. 49y² – 16x² = 784

Ans.

$\begin{array}{l} Comparing \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 with \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, we get \\ a^{2} = 16 and b^{2} = 9 \\ \Rightarrow a = \pm 4 and b = \pm 3 \\ c = \pm \sqrt{a^{2} + b^{2}} \\ = \pm \sqrt{16 + 9} \\ = \pm \sqrt{25} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} c = \pm 5 \\ Coordinates of foci are (\pm 5, 0) . \\ Coordinates of vertices are (\pm 4, 0) . \\ Eccentricity = \frac{c}{a} \\ = \frac{5}{4} \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 {(3)}^{2}}{4} \\ = \frac{9}{2} \end{array}$

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

$\begin{array}{l} Coordinates of vertices are (0, \pm 3) . \\ Eccentricity = \frac{c}{a} \\ = \frac{6}{3} = 2 \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 \times 27}{3} \\ = 18 \end{array}$

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

$\begin{array}{l} Coordinates of vertices are (0, \pm 2) . \\ Eccentricity = \frac{c}{a} \\ = \frac{\sqrt{13}}{2} \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 \times 9}{2} \\ = 9 \end{array}$

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

$\begin{array}{l} Coordinates of vertices are (\pm 6, 0) . \\ Eccentricity = \frac{c}{a} \\ = \frac{10}{6} \\ = \frac{5}{3} \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 {(8)}^{2}}{6} \\ = \frac{64}{3} \end{array}$

$\begin{array}{l} The given equation of hyperbola is: \\ 5 y^{2} - 9 x^{2} = 36 \\ \Rightarrow \frac{{5y}^{2}}{36} - \frac{{9x}^{2}}{36} = \frac{36}{36} \\ \Rightarrow \frac{y^{2}}{(\frac{36}{5})} - \frac{x^{2}}{4} = 1 . .. (i) \\ Comparing \frac{y^{2}}{(\frac{36}{5})} - \frac{x^{2}}{4} = 1 with \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1, we get \\ a^{2} = \frac{36}{5} {and b}^{2} = 4 \end{array}$

$\begin{array}{l} \Rightarrow a = \pm \frac{6}{\sqrt{5}} and b = \pm 2 \\ c = \pm \sqrt{a^{2} + b^{2}} \\ = \pm \sqrt{\frac{36}{5} + 4} \\ c = \pm \sqrt{\frac{36 + 20}{5}} \\ = \pm \sqrt{\frac{56}{5}} = \pm \frac{2 \sqrt{14}}{\sqrt{5}} \\ Coordinates of foci are (0, \pm \frac{2 \sqrt{14}}{\sqrt{5}}) . \\ Coordinates of vertices are (0, \pm \frac{6}{\sqrt{5}}) . \\ Eccentricity = \frac{c}{a} \\ = \frac{(\frac{2 \sqrt{14}}{\sqrt{5}})}{(\frac{6}{\sqrt{5}})} \\ = \frac{\sqrt{14}}{3} \\ Length of latus rectum = \frac{2 b^{2}}{a} \\ = \frac{2 \times 2^{2}}{(\frac{6}{\sqrt{5}})} \end{array}$ $Length of latus rectum = \frac{4 \sqrt{5}}{3}$

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

$\begin{array}{l} = \frac{2 \times 7^{2}}{4} \end{array}$

$\begin{array}{l} Length of latus rectum = \frac{49}{2} \end{array}$

Q.2 Find the equation of the hyperbola satisfying the below condition:

$Vertices (\pm 2, 0), foci (\pm 3, 0)$

Ans.

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

Q.3 Find the equation of the hyperbola satisfying the below condition:

$Vertices (0, \pm 5), foci (0, \pm 8)$

Ans.

$Since, foci is on y-axis, the equation of hyperbola is of the form$

$\begin{array}{l} \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \\ Since, vertices (0, \pm 5), foci (0, \pm 8) \\ So, a = 5 and c = 8 . \\ ∵ b^{2} = c^{2} - a^{2} \\ \Rightarrow b^{2} = 8^{2} - 5^{2} \\ = 64 - 25 \\ = 39 \\ \Rightarrow b = \sqrt{39} \\ Therefore, equation of hyperbola is \\ \frac{y^{2}}{5^{2}} - \frac{x^{2}}{{(\sqrt{39})}^{2}} = 1 \\ \Rightarrow \frac{y^{2}}{25} - \frac{x^{2}}{39} = 1 \end{array}$

Q.4 Find the equation of the hyperbola satisfying the below condition:

$Vertices (0, \pm 3), foci (0, \pm 5)$

Ans.

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

$\begin{array}{l} Therefore, equation of hyperbola is \\ \frac{y^{2}}{3^{2}} - \frac{x^{2}}{4^{2}} = 1 \\ \Rightarrow \frac{y^{2}}{9} - \frac{x^{2}}{16} = 1 \end{array}$

Q.5 Find the equation of the hyperbola satisfying the below condition:

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

Ans.

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

Q.6 Find the equation of the hyperbola satisfying the below condition:

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

Ans.

$\begin{array}{l} Foci (0, \pm 13), the conjugate axis is of length 24 . \\ Since, foci is on y-axis, the equation of hyperbola is of \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} the form \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \\ Here, c = 13 and b = \frac{24}{2} = 12 \\ ∵ a^{2} = c^{2} - b^{2} \\ = 13^{2} - {(12)}^{2} \\ = 169 - 144 \\ = 25 \\ \Rightarrow a = \pm 5 \\ Therefore, the equation of hyperbola is \\ \frac{y^{2}}{5^{2}} - \frac{x^{2}}{{(12)}^{2}} = 1 \\ \frac{y^{2}}{25} - \frac{x^{2}}{144} = 1 \end{array}$

Q.7 Find the equation of the hyperbola satisfying the below condition:

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

Ans.

$\begin{array}{l} Given : Foci (\pm 3 \sqrt{5}, 0), the latus rectum is of length 8 . \\ Since, foci is on x-axis, the equation of hyperbola is of \\ the form \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \\ Here, c = 3 \sqrt{5} and \frac{{2b}^{2}}{a} = 8 \\ \Rightarrow b^{2} = 4 a \\ ∵ b^{2} = c^{2} - a^{2} \\ 4 a = {(3 \sqrt{5})}^{2} - a^{2} \end{array}$

$\begin{array}{l} 4 a = 45 - a^{2} \\ a^{2} + 4 a - 45 = 0 \\ \Rightarrow (a + 9) (a - 5) = 0 \\ \Rightarrow a = 5, - 9 \\ \Rightarrow a = 5 (Neglecting - 9) \\ and b^{2} = 4 \times 5 \\ = 20 \\ Therefore, the equation of hyperbola is \\ \frac{x^{2}}{5^{2}} - \frac{y^{2}}{20} = 1 \\ \frac{x^{2}}{25} - \frac{y^{2}}{20} = 1 \end{array}$

Q.8 Find the equation of the hyperbola satisfying the below condition:

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

Ans.

$\begin{array}{l} Given : Foci (\pm 4, 0), the latus rectum is of length 12 . \\ Since, foci is on x-axis, the equation of hyperbola is of \\ the form \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \\ Here, c = 4 and \frac{{2b}^{2}}{a} = 12 \\ \Rightarrow b^{2} = 6 a \\ ∵ b^{2} = c^{2} - a^{2} \\ 6 a = 4^{2} - a^{2} \\ 6 a = 16 - a^{2} \\ a^{2} + 6 a - 16 = 0 \\ \Rightarrow (a + 8) (a - 2) = 0 \end{array}$

$\begin{array}{l} \Rightarrow a = 2, - 8 \\ \Rightarrow a = 2 (Neglecting - 8) \\ and b^{2} = 6 \times 2 [∵ b^{2} = 6 a] \\ = 12 \\ Therefore, the equation of hyperbola is \\ \frac{x^{2}}{2^{2}} - \frac{y^{2}}{12} = 1 \\ \frac{x^{2}}{4} - \frac{y^{2}}{12} = 1 \end{array}$

Q.9 Find the equation of the hyperbola satisfying the below condition:

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

Ans.

$\begin{array}{l} Given : Vertices = (\pm 7, 0), e = \frac{4}{3} \\ Since, vertices lie on x-axis, the equation of hyperbola is \\ of the form \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \\ Here, a = 7 and e = \frac{4}{3} \\ ∵ e = \frac{c}{a} \\ \Rightarrow c = ae = 7 \times \frac{4}{3} = \frac{28}{3} \\ and b^{2} = c^{2} - a^{2} \\ b^{2} = {(\frac{28}{3})}^{2} - 7^{2} \end{array}$

$\begin{array}{l} = \frac{784}{9} - 49 \\ = \frac{784 - 441}{9} \\ = \frac{343}{9} \\ Therefore, the equation of hyperbola is \\ \frac{x^{2}}{7^{2}} - \frac{y^{2}}{\frac{343}{9}} = 1 \\ \Rightarrow \frac{x^{2}}{49} - \frac{9 y^{2}}{343} = 1 \end{array}$

Q.10 Find the equation of the hyperbola satisfying the below condition:

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

Ans.

$\begin{array}{l} Foci (0, \pm \sqrt{10}), passing through (2, 3) . \\ Since, foci is on y-axis, the equation of hyperbola is of \\ the form \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \\ Here, c = \sqrt{10} \\ ∵ c^{2} = a^{2} + b^{2} \\ \Rightarrow 10 = a^{2} + b^{2} . .. (i) \\ Since, hyperbola passes through the point (2, 3) . \\ So, \frac{3^{2}}{a^{2}} - \frac{2^{2}}{b^{2}} = 1 \\ \frac{9}{a^{2}} - \frac{4}{b^{2}} = 1 \end{array}$

$\begin{array}{l} Putting b^{2} = 10 - a^{2}, from equation (i), we get \\ \frac{9}{a^{2}} - \frac{4}{10 - a^{2}} = 1 \\ \Rightarrow \frac{9 (10 - a^{2}) - 4 a^{2}}{a^{2} (10 - a^{2})} = 1 \\ \Rightarrow 90 - 9 a^{2} - 4 a^{2} = 10 a^{2} - a^{4} \\ \Rightarrow a^{4} - 23 a^{2} + 90 = 0 \\ \Rightarrow (a^{2} - 18) (a^{2} - 5) = 0 \\ \Rightarrow a^{2} = 18, 5 \\ So, b^{2} = 10 - a^{2} \\ = 10 - 5 \\ = 5 \\ Neglecting a^{2} = 18 because b^{2} = 10 - a^{2} . \\ Therefore, the equation of required hyperbola is \\ \frac{y^{2}}{5} - \frac{x^{2}}{5} = 1 \end{array}$

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The NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 are enough for helping students strengthen their fundamental ideas and get ready to manage any problems related to them. Students can practice past papers and sample papers for better preparation. Students who fully grasp the ideas score better in the examination by practising the NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 accessible on the Extramarks’ website.

2. Are the NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 available on the Extramarks website?

The NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 are available on the Extramarks website as well as on the mobile application. Students may discover Class 11 Maths Chapter 11 Exercise 11.4. Additionally, it provides K12 study resources for the students’ tests together with live sessions with experts who can thoroughly address all of their questions. The learning software also provides extensive coverage of the syllabus, gamified learning opportunities, curriculum mapping, and much more so that students can learn in a planned and interesting way.

3. Do students find the exercises related to the NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 difficult?

Students can find the textbook exercises tough. They can simply practice the NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 and excel in the crucial topics to do well in their examinations with consistent practice and the appropriate direction from Extramarks. The NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 are provided in-depth by Extramarks to help students comprehend the concepts in the chapter. It is also possible for students who are reluctant to ask questions in front of their peers at school to participate in doubt sessions through Extramarks.

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.4) Exercise 11.4

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.4) Exercise 11.4

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.4

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FAQs (Frequently Asked Questions)

1. Are the NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 enough to prepare for the examination?

2. Are the NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 available on the Extramarks website?

3. Do students find the exercises related to the NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 difficult?

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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.4) Exercise 11.4

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.4) Exercise 11.4

Access NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.4

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FAQs (Frequently Asked Questions)

1. Are the NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 enough to prepare for the examination?

2. Are the NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 available on the Extramarks website?

3. Do students find the exercises related to the NCERT Solutions For Class 11 Maths Chapter 11 Exercise 11.4 difficult?

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NCERT Solutions Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions