Question

Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$ If the norm al at the point $P$ intersects the $X$ -axis at $(9, 0)$, then the eccentricity of the hyperbola is

• A. $\sqrt\frac{5}{2}$
• B. $\sqrt\frac{3}{2}$
• C. $\sqrt2$
• D. $\sqrt3$

Answer 1

Answer: (B)

Equation of normal to hyperbola at $\left(x_{1}, y_{1}\right)$ is
$ \frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}=\left(a^{2}+b^{2}\right) $
$\therefore $ At $(6,3)=\frac{a^{2} x}{6}+\frac{b^{2} y}{3}=\left(a^{2}+b^{2}\right)$
$\because $ It passes through $(9,0) . \Rightarrow \frac{a^{2} \cdot 9}{6}=a^{2}+b^{2}$
$\Rightarrow \frac{3 a^{2}}{2}-a^{2}=b^{2} \Rightarrow \frac{a^{2}}{b^{2}}=2$
$\therefore e^{2}=1+\frac{b^{2}}{a^{2}}=1+\frac{1}{2} \Rightarrow e=\sqrt{\frac{3}{2}}$