Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at the point $P$…
Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at the point $P$ intersects the $X$-axis at $(9,0)$, then the eccentricity of the hyperbola is
$\sqrt{\frac{5}{2}}$
$\sqrt{\frac{3}{2}}$
$\sqrt{2}$
$\sqrt{3}$
Solution
Equation of normal to hyperbola at $\left(x_1, y_1\right)$ is
$
\begin{gathered}
\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2 \\
\therefore \text { At }(6,3), \frac{a^2 x}{6}+\frac{b^2 y}{3}=a^2+b^2
\end{gathered}
$
It passes throught $(9,0)$.
$
\begin{aligned}
& \text { Now, } \quad \frac{a^2 \cdot 9}{6}=a^2+b^2 \\
& \Rightarrow \quad \frac{3 a^2}{2}-a^2=b^2 \Rightarrow \frac{a^2}{b^2}=2 \\
& \therefore \quad e^2=1+\frac{b^2}{a^2}=1+\frac{1}{2} \Rightarrow e=\sqrt{\frac{3}{2}}
\end{aligned}
$