Question

The tangent at an extremity (in the first quadrant) of latus rectrum of the hyperbola $\frac{x^2}{4}-\frac{y^2}{5}=1$, meets x-axis and y-axis at $A$ and $B$ respectively. Then $(OA{)}^2-(OB{)}^2$, where $O$ is the origin, equals:

• A. $-\frac{20}{9}$
• B. $\frac{16}{9}$
• C. $4$
• D. $-\frac{4}{3}$

Answer 1

The correct option is A $-\frac{20}{9}$

Equation of hyperbola will be $\frac{x^2}{4}-\frac{y^2}{5}=1$

So, $a^2=4$ and $b^2=5$

We know that $e^2{a}^2=a^2+b^2$

$4e^2=4+5$

$e^2=\frac{9}{4}$

$e=\frac{3}{2}$

Latusrectum L will be $(ae,\frac{b^2}{a})$

$L=(2\times \frac{3}{2},\frac{5}{2})=(3,\frac{5}{2})$

We know that equation of tangent $(x_1,y_1)$ will be $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$

$\frac{3x}{4}-\frac{y}{2}=1$

X-intercept will be $\frac{4}{3}$

Y-intercept will be $-2$

$O{A}^2-O{B}^2=\frac{16}{9}-4=\frac{-20}{9}$