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. $\sqrt{2}$
• D. $\sqrt{3}$

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 $(9,0).\Rightarrow \frac{a^2\cdot 9}{6}=a^2+b^2$
$\Rightarrow \frac{3a^2}{2}-a^2=b^2\Rightarrow \frac{a^2}{b^2}=2$
$\therefore e^2=1+\frac{b^2}{a^2}=1+\frac{1}{2}\Rightarrow e=\sqrt{\frac{3}{2}}$