Question

The latus rectum of the hyberbola $9x^2-16y^2-18x-32y-151=0$ is

• A. $\frac{9}{2}$
• B. 9
• C. $\frac{3}{2}$
• D. $\frac{9}{4}$

Answer 1

Answer: (A)

Hyperbola $9x^{2}-16y^{2}-18x-32y-151 = 0$ can be written as
$9\left(x^{2}-2x\right)-16\left(y^{2}+2y\right) = 151 $
$ \Rightarrow 9\left(x-1\right)^{2}-16\left(y+1\right)^{2}=151+9-16 $
$ = 144 $
$ \Rightarrow \frac{\left(x-1\right)^{2}}{6} - \frac{\left(y+1\right)^{2}}{9} = 1 $
$ \Rightarrow \frac{X^{2}}{16}-\frac{Y^{2}}{9}=1 $
$ \quad\left(X= x-1, Y= y-1\right)$
Here $a^{2}= 16, b^{2} = 9$
Latus Rectum $= 2 \frac{b^{2}}{a} =\frac{ 2\times9}{4} = \frac{9}{2}$