Question

If a normal drawn at one end of the latus rectum of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the axes at points $A\&B$ respectively, then area of $△OAB$ (in sq.units) is

• A. $a^2{e}^5$
• B. $\frac{a^2{e}^5}{2}$
• C. $\frac{a^2{e}^5}{4}$
• D. $\frac{a^2{e}^5}{8}$

Answer 1

The correct option is B $\frac{a^2{e}^5}{2}$

Let one end of the latus rectum of the given hyperbola is $L(ae,\frac{b^2}{a})$

The equation of the normal to the given hyperbola at $L$ is
$\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$
$\Rightarrow \frac{ax}{e}+ay=a^2+b^2$
$\Rightarrow \frac{x}{\left(\frac{e(a^2+b^2)}{a}\right)}+\frac{y}{\left(\frac{a^2+b^2}{a}\right)}=1$
Hence, the area of $△OAB$
$=\frac{1}{2}|\times \frac{e(a^2+b^2)}{a}\times \left(\frac{a^2+b^2}{a}\right)|$
$=\frac{1}{2}\times \frac{e(a^2+b^2{)}^2}{a^2}$
$=\frac{1}{2}\times a^2{e}^5$ sq. units