Question

Let $P(8,4)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at point $P$ intersects the $x-$axis at $(12,0)$, then the value of eccentricity is

• A. $\sqrt{\frac{7}{2}}$
• B. $\sqrt{\frac{3}{2}}$
• C. $\sqrt{5}$
• D. $\sqrt{3}$

Answer 1

The correct option is B $\sqrt{\frac{3}{2}}$

Equation of normal at $(x_1,y_1)$ is
$\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$
Equation of normal at $(8,4)$ is
$\frac{a^{2}x}{8}+\frac{b^{2}y}{4}=a^2+b^2$
This equation passes through $(12,0)$
Thus $\frac{a^2\left(12\right)}{8}+0=a^2+b^2$
$\Rightarrow \frac{b^2}{a^2}=\frac{1}{2}\cdots \left(1\right)$
Now $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$