MATCH THE FOLLOWING
Hyperbola Length of latusrectum
A} $x^{2} - 4 y^{2} = 4$ 1. 1
B} $25 x^{2} - 16 y^{2} = 400$ 2.12
C} $2 x^{2} - y^{2} - 4 x - 4 y - 20 = 0$ 3.9/2
D) $9 x^{2} - 16 y^{2} + 72 x - 32 y - 16 = 0$ 4. 25/2
The correct match is

I II III IV
1 2 3 4

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1 4 2 3

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3 1 2 4

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2 3 4 1

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Solution

The correct option is A 1 4 2 3
(A) $\frac{x^{2}}{4} - y^{2} = 1$

$L R = \frac{2 b^{2}}{a}$

$= \frac{2}{2}$

$= 1$

(B) $\frac{x^{2}}{16} - \frac{y^{2}}{25} = 1$

$b = 5, a = 4$

$L R = \frac{2.25}{4}$

$= \frac{25}{2}$

(C) $2 (x - 1)^{2} - (y - 2)^{2} = 18$

$\frac{(x - 1)^{2}}{9} - \frac{(y - 2)^{2}}{18} = 1$

$b^{2} = 18, a = 3$

$L R = \frac{2 \times 18}{3}$

$= 12$

(D) $\frac{(x + 4)^{2}}{16} - \frac{(y + 1)^{2}}{9} = 1$

$b^{2} = 9$

$a = 4$

$L R = 2 \times \frac{9}{4}$

$= \frac{9}{2}$

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The lengths of the axes of the hypberbola $9 x^{2} - 16 y^{2} + 72 x - 32 y - 16$ = 0 are

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