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
$9\left(x^2-2x\right)-16\left(y^2+2y\right)=151$
$\Rightarrow 9{\left(x-1\right)}^2-16{\left(y+1\right)}^2=151+9-16$
$=144$
$\Rightarrow \frac{{\left(x-1\right)}^2}{6}-\frac{{\left(y+1\right)}^2}{9}=1$
$\Rightarrow \frac{X^2}{16}-\frac{Y^2}{9}=1$
$\left(X=x-1,Y=y-1\right)$
Here $a^2=16,b^2=9$
Latus Rectum $=2\frac{b^2}{a}=\frac{2\times 9}{4}=\frac{9}{2}$