Question

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

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

Answer 1

The correct option is D $\frac{9}{2}$

Given the equation of the hyperbola is,

$9x^2-16y^2-18x-32y-151=0$

or, $9(x^2-2x)-16(y^2-2y)=151$

or, $9(x-1{)}^2-16(y-1{)}^2=144$

or, $\frac{(x-1{)}^2}{16}+\frac{(y-1{)}^2}{9}=1$.

Comparing this equation with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ we get,

$a^2=16,b^2=9$.

So length of latus rectum $=2\times \frac{b^2}{a}=\frac{2\times 9}{4}=\frac{9}{2}$ units.