Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.4

In each of the following questions 1 to 6, find the co-ordinates of the foci and the vertices, the eccentricity and length of latus rectum of the hyperbola:

1. \(\frac{x^{2}}{16}\) – \(\frac{y^{2}}{9}\) = 1

2. \(\frac{y^{2}}{9}\) – \(\frac{x^{2}}{27}\) = 1

3. 9y^{2} – 4x^{2} = 36

4. 16x^{2} – 9y^{2} = 576

5. 5y^{2} – 9x^{2} = 36

6. 49y^{2} – 16x^{2} = 784

Solutions to questions 1-6:

1. Equation of the hyperbola is

\(\frac{x^{2}}{16}\) – \(\frac{y^{2}}{9}\) = 1

Here, a^{2} = 16 â‡’ a = 4, b^{2} = 9 â‡’ b = 3.

âˆ´ c^{2} = a^{2} + b^{2} = 16 + 9 = 25.

âˆ´ c = 5.

Axis is x-axis.

Co-ordinates of foci (Â± 5, 0).

Co-ordinates of vertices (Â± 4, 0).

Eccentricity e = \(\frac{c}{a}\) = \(\frac{5}{4}\).

Latus rectum = \(\frac{2b^{2}}{a}\) = \(\frac{2Ã—9}{4}\) = \(\frac{9}{2}\).

2. Equation of hyperbola is \(\frac{y^{2}}{9}\) – \(\frac{x^{2}}{27}\) = 1.

Here, a^{2} = 9, b^{2} = 27,

â‡’ c^{2} = a^{2} + b^{2}

= 9 + 27 = 36.

a = 3,b = 3\(\sqrt{3}\), c = 6.

Axis of hyperbola is y-axis.

âˆ´ Foci are 5(0, Â± 6).

Vertices (0, Â± 3)

Eccentricity e = \(\frac{c}{a}\) = \(\frac{6}{3}\) = 2.

Latus rectum = \(\frac{2b^{2}}{a}\) = \(\frac{2Ã—27}{3}\) = 18.

3. 9y^{2} – 4x^{2} = 36 is the equation of hyperbola.

i.e; \(\frac{y^{2}}{4}\) – \(\frac{x^{2}}{9}\) = 1.

a^{2} = 4, b^{2} = 9, c^{2} = a^{2} + b^{2} = 4 + 9 = 13.

âˆ´ a = 2, b = 3, c = \(\sqrt{3}\).

Axis is y-axis.

Foci (0, Â± \(\sqrt{13}\)), vertices (0, Â± 2).

Eccentricity e = \(\frac{c}{a}\) = \(\frac{10}{6}\) = \(\frac{5}{3}\).

Latus rectum = \(\frac{2b^{2}}{a}\) = \(\frac{2Ã—64}{6}\) = \(\frac{64}{3}\).

5. Equation of hyperbola is 5y^{2} – 9x^{2} = 36.

Axis is along y-axis.

6. Equation of hyperbola is 49y^{2} – 16x^{2} = 784.

or \(\frac{y^{2}}{16}\) – \(\frac{x^{2}}{49}\) = 1.

âˆ´ a^{2} = 16, b^{2} = 49, c^{2} = 16 + 49 = 65.

â‡’ a = 4, 5 = 7, c = \(\sqrt{65}\).

Axis is y-axis.

Foci are (0, Â± \(\sqrt{65}\)) and vertices are (0, Â± 4).

Eccentricity e = \(\frac{c}{a}\) = \(\frac{\sqrt{65}}{4}\),

latus rectum = \(\frac{2b^{2}}{a}\) = \(\frac{2Ã—49}{4}\) = \(\frac{49}{2}\).

In each of the following questions 7 to 15, find the equations of hyperbola satisfying the given conditions:

7. Vertices (Â± 2,0), foci (Â± 3, 0)

8. Vertices (0, Â± 5), foci (0, Â± 8)

9. Vertices (0, Â± 3), foci (0, Â± 5)

10. Foci (Â± 5, 0), the transverse axis is of length 8.

11. Foci (0, Â± 13), the conjugate axis is of length 24.

12. Foci (Â± 3\(\sqrt{5}\), 0), the latus rectum is length 8.

13. Foci (Â± 4, 0), the latus rectum is of length 12.

14. Vertices (Â± 4, 0), e = \(\frac{4}{3}\).

15. Foci (0, Â± \(\sqrt{10}\)), passing through (2, 3).

Solutions to questions 7-15:

7. We have: vertices (Â± 2, 0) and foci (Â± 3, 0).

Axis of hyperbola is x-axis.

a = 2, c = 3 â‡’ b^{2} = c^{2} – a^{2} = 9 – 4 = 5.

âˆ´ b = \(\sqrt{5}\).

Equation of hyperbola is

\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{5}\) = 1.

8. The given vertices are (0, Â± 5), which lie on y-axis.

Thus, equation of the hyperbola is of the form

\(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1

The vertices of hyperbola (1) are (0, Â± a). Here, the vertices are given to be (0, Â± 5).

âˆ´ a = 5.

The co-ordinates of foci of (1) are (0, Â± ae). Here, they are given to be (0, Â± 8).

So, ae = 8 â‡’ 5e = 8 â‡’ e = \(\frac{8}{5}\).

Hence, the required equation of the hyperbola is

9. Vertices are (0, Â± 3) â‡’ a = 3.

Foci are (0, Â± 5) â‡’ c = 5.

So, b^{2} = c^{2} – a^{2}

= 25 – 9 = 16

â‡’ b = 4.

âˆ´ Equation of hyperbola is (axis being y-axis):

\(\frac{y^{2}}{9}\) – \(\frac{x^{2}}{16}\) = 1.

10. Foci are (Â± 5, 0) â‡’ c = 5.

Transverse axis = 8 â‡’ a = 4.

âˆ´ b^{2} = c^{2} – a^{2}

= 25 – 16 = 9

â‡’ b = 3.

Axis of hyperbola is x-axis, since (Â± 5, 0) lies on it. Therefore, equation of the hyperbola is

\(\frac{x^{2}}{16}\) – \(\frac{y^{2}}{9}\) = 1.

11. Foci are (0, Â± 13) â‡’ c = 13.

The length of conjugate axis = 26 = 24 â‡’ b = 12.

a^{2} = c^{2} – b^{2}

= 169 – 144

= 25

â‡’ a = 5

Axis of hyperbola is y-axis, since (0, Â± 13) lies on it.

\(\frac{y^{2}}{25}\) – \(\frac{x^{2}}{144}\) = 1.

12. Foci are (Â± 3\(\sqrt{5}\), 0) â‡’ c = 3\(\sqrt{5}\).

The length of latus rectum = \(\frac{2b^{2}}{a}\) = 8.

âˆ´ b^{2} = 4a ……………… (1)

Also, c^{2} = a^{2} + b^{2} i.e., 45 = a^{2} + b^{2} …………………. (2)

Eliminating, from (1) and (2), we get

45 = a^{2} + 4a or a^{2} + 4a – 45 = 0.

or (a + 9)(a – 5) = 0.

a â‰ – 9 âˆ´a = 5 â‡’ b^{2} = 4a = 4 Ã— 5 = 20.

a^{2} = 25 and b^{2} = 20

âˆ´ Equation of hyperbola is [Since (Â± 3\(\sqrt{5}\), 0) lies on x-axis]

\(\frac{x^{2}}{25}\) – \(\frac{y^{2}}{20}\) = 1.

13. Foci are (Â± 4, 0) â‡’ c = 4

or c^{2} = a^{2} + b^{2} â‡’ 16 = a^{2} + b^{2} ……………. (1)

âˆ´ Latus rectum = \(\frac{2b^{2}}{a}\) = 12.

âˆ´ b^{2} = 6a ……………. (2)

Eliminating b^{2} from (1) and (2), we get

âˆ´ 16 = a^{2} + 6a or a^{2} + 6a – 16 = 0.

or (a + 8)(a – 2) = 0 [âˆµa â‰ – 8]

âˆ´ a = 2 â‡’ b^{2} = 6a = 6 Ã— 2 = 12.

âˆ´ a^{2} = 4, b = 12, Axis is x-axis.

So, equation of the hyperbola is

\(\frac{x^{2}}{4}\) – \(\frac{y^{2}}{12}\) = 1.

14. Vertices are (Â± 7, 0) â‡’ a = 7.

Axis is along x-axis.

So, equation of the hyperbola is

\(\frac{x^{2}}{49}\) – \(\frac{y^{2}}{343/9}\) = 1 or 7x^{2} – 9y^{2} = 343.

15. The given foci are (0, Â± \(\sqrt{10}\)), which lies on y-axis.

Let the equation of hyperbola be

\(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1.

âˆ´ ae = \(\sqrt{10}\).

Also, b^{2} = a^{2}(e^{2} – 1) = a^{2}e^{2} – a^{2}.

= 10 – a^{2}. ………………….. (2)

Thus, the equation of the hyperbola is

\(\frac{y^{2}}{a^{2}}\) – \(\frac{x^{2}}{10-a^{2}}\) = 1.

As it passes through the point (2, 3), so

= \(\frac{9}{a^{2}}\) – \(\frac{4}{10-a^{2}}\) = 1 â‡’ 90 – 9a^{2} – 4a^{2} = a^{2}(10 – a^{2})

â‡’ a^{4} – 23a^{2} + 90 = 0 â‡’ (a^{2} – 5)(a^{2} – 18) = 0.

â‡’ a^{2} = 18, 5.

When a^{2} = 18, then from (2), b^{2} = – 8, which is not possible and when a^{2} = 5, then from (2), b^{2} = 5.

Hence, the required equation of the hyperbola is

\(\frac{y^{2}}{5}\) = \(\frac{x^{2}}{5}\) = 1 â‡’ y^{2} – x^{2} = 5.