# GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.4

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. 9y2 – 4x2 = 36
4. 16x2 – 9y2 = 576
5. 5y2 – 9x2 = 36
6. 49y2 – 16x2 = 784
Solutions to questions 1-6:
1. Equation of the hyperbola is
$$\frac{x^{2}}{16}$$ – $$\frac{y^{2}}{9}$$ = 1
Here, a2 = 16 ⇒ a = 4, b2 = 9 ⇒ b = 3.
∴ c2 = a2 + b2 = 16 + 9 = 25.
∴ c = 5.
Axis is x-axis.
Co-ordinates of foci (± 5, 0).
Co-ordinates of vertices (± 4, 0).
Eccentricity e = $$\frac{c}{a}$$ = $$\frac{5}{4}$$.
Latus rectum = $$\frac{2b^{2}}{a}$$ = $$\frac{2×9}{4}$$ = $$\frac{9}{2}$$.

2. Equation of hyperbola is $$\frac{y^{2}}{9}$$ – $$\frac{x^{2}}{27}$$ = 1.
Here, a2 = 9, b2 = 27,
⇒ c2 = a2 + b2
= 9 + 27 = 36.
a = 3,b = 3$$\sqrt{3}$$, c = 6.
Axis of hyperbola is y-axis.
∴ Foci are 5(0, ± 6).
Vertices (0, ± 3)
Eccentricity e = $$\frac{c}{a}$$ = $$\frac{6}{3}$$ = 2.
Latus rectum = $$\frac{2b^{2}}{a}$$ = $$\frac{2×27}{3}$$ = 18.

3. 9y2 – 4x2 = 36 is the equation of hyperbola.
i.e; $$\frac{y^{2}}{4}$$ – $$\frac{x^{2}}{9}$$ = 1.
a2 = 4, b2 = 9, c2 = a2 + b2 = 4 + 9 = 13.
∴ a = 2, b = 3, c = $$\sqrt{3}$$.
Axis is y-axis.
Foci (0, ± $$\sqrt{13}$$), vertices (0, ± 2).
Eccentricity e = $$\frac{c}{a}$$ = $$\frac{10}{6}$$ = $$\frac{5}{3}$$.
Latus rectum = $$\frac{2b^{2}}{a}$$ = $$\frac{2×64}{6}$$ = $$\frac{64}{3}$$.

5. Equation of hyperbola is 5y2 – 9x2 = 36.

Axis is along y-axis.

6. Equation of hyperbola is 49y2 – 16x2 = 784.
or $$\frac{y^{2}}{16}$$ – $$\frac{x^{2}}{49}$$ = 1.
∴ a2 = 16, b2 = 49, c2 = 16 + 49 = 65.
⇒ a = 4, 5 = 7, c = $$\sqrt{65}$$.
Axis is y-axis.
Foci are (0, ± $$\sqrt{65}$$) and vertices are (0, ± 4).
Eccentricity e = $$\frac{c}{a}$$ = $$\frac{\sqrt{65}}{4}$$,
latus rectum = $$\frac{2b^{2}}{a}$$ = $$\frac{2×49}{4}$$ = $$\frac{49}{2}$$.

In each of the following questions 7 to 15, find the equations of hyperbola satisfying the given conditions:
7. Vertices (± 2,0), foci (± 3, 0)
8. Vertices (0, ± 5), foci (0, ± 8)
9. Vertices (0, ± 3), foci (0, ± 5)
10. Foci (± 5, 0), the transverse axis is of length 8.
11. Foci (0, ± 13), the conjugate axis is of length 24.
12. Foci (± 3$$\sqrt{5}$$, 0), the latus rectum is length 8.
13. Foci (± 4, 0), the latus rectum is of length 12.
14. Vertices (± 4, 0), e = $$\frac{4}{3}$$.
15. Foci (0, ± $$\sqrt{10}$$), passing through (2, 3).
Solutions to questions 7-15:
7. We have: vertices (± 2, 0) and foci (± 3, 0).
Axis of hyperbola is x-axis.
a = 2, c = 3 ⇒ b2 = c2 – a2 = 9 – 4 = 5.
∴ b = $$\sqrt{5}$$.
Equation of hyperbola is
$$\frac{x^{2}}{4}$$ – $$\frac{y^{2}}{5}$$ = 1.

8. The given vertices are (0, ± 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, ± a). Here, the vertices are given to be (0, ± 5).
∴ a = 5.
The co-ordinates of foci of (1) are (0, ± ae). Here, they are given to be (0, ± 8).
So, ae = 8 ⇒ 5e = 8 ⇒ e = $$\frac{8}{5}$$.

Hence, the required equation of the hyperbola is

9. Vertices are (0, ± 3) ⇒ a = 3.
Foci are (0, ± 5) ⇒ c = 5.
So, b2 = c2 – a2
= 25 – 9 = 16
⇒ b = 4.
∴ Equation of hyperbola is (axis being y-axis):
$$\frac{y^{2}}{9}$$ – $$\frac{x^{2}}{16}$$ = 1.

10. Foci are (± 5, 0) ⇒ c = 5.
Transverse axis = 8 ⇒ a = 4.
∴ b2 = c2 – a2
= 25 – 16 = 9
⇒ b = 3.
Axis of hyperbola is x-axis, since (± 5, 0) lies on it. Therefore, equation of the hyperbola is
$$\frac{x^{2}}{16}$$ – $$\frac{y^{2}}{9}$$ = 1.

11. Foci are (0, ± 13) ⇒ c = 13.
The length of conjugate axis = 26 = 24 ⇒ b = 12.
a2 = c2 – b2
= 169 – 144
= 25
⇒ a = 5
Axis of hyperbola is y-axis, since (0, ± 13) lies on it.
$$\frac{y^{2}}{25}$$ – $$\frac{x^{2}}{144}$$ = 1.

12. Foci are (± 3$$\sqrt{5}$$, 0) ⇒ c = 3$$\sqrt{5}$$.
The length of latus rectum = $$\frac{2b^{2}}{a}$$ = 8.
∴ b2 = 4a ……………… (1)
Also, c2 = a2 + b2 i.e., 45 = a2 + b2 …………………. (2)
Eliminating, from (1) and (2), we get
45 = a2 + 4a or a2 + 4a – 45 = 0.
or (a + 9)(a – 5) = 0.
a ≠ – 9 ∴a = 5 ⇒ b2 = 4a = 4 × 5 = 20.
a2 = 25 and b2 = 20
∴ Equation of hyperbola is [Since (± 3$$\sqrt{5}$$, 0) lies on x-axis]
$$\frac{x^{2}}{25}$$ – $$\frac{y^{2}}{20}$$ = 1.

13. Foci are (± 4, 0) ⇒ c = 4
or c2 = a2 + b2 ⇒ 16 = a2 + b2 ……………. (1)
∴ Latus rectum = $$\frac{2b^{2}}{a}$$ = 12.
∴ b2 = 6a ……………. (2)
Eliminating b2 from (1) and (2), we get
∴ 16 = a2 + 6a or a2 + 6a – 16 = 0.
or (a + 8)(a – 2) = 0 [∵a ≠ – 8]
∴ a = 2 ⇒ b2 = 6a = 6 × 2 = 12.
∴ a2 = 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 (± 7, 0) ⇒ a = 7.

Axis is along x-axis.
So, equation of the hyperbola is
$$\frac{x^{2}}{49}$$ – $$\frac{y^{2}}{343/9}$$ = 1 or 7x2 – 9y2 = 343.

15. The given foci are (0, ± $$\sqrt{10}$$), which lies on y-axis.
Let the equation of hyperbola be
$$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}$$ = 1.
∴ ae = $$\sqrt{10}$$.
Also, b2 = a2(e2 – 1) = a2e2 – a2.
= 10 – a2. ………………….. (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 ⇒ 90 – 9a2 – 4a2 = a2(10 – a2)
⇒ a4 – 23a2 + 90 = 0 ⇒ (a2 – 5)(a2 – 18) = 0.
⇒ a2 = 18, 5.
When a2 = 18, then from (2), b2 = – 8, which is not possible and when a2 = 5, then from (2), b2 = 5.
Hence, the required equation of the hyperbola is
$$\frac{y^{2}}{5}$$ = $$\frac{x^{2}}{5}$$ = 1 ⇒ y2 – x2 = 5.