JEE Advanced 2011 Paper 2Question 1
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
Hyperbola
Solution
Normal slope at $(6,3)$ is $-\frac{3a^2}{6b^2}$; the line to $(9,0)$ has slope $-1$, so $a^2=2b^2$. Then $e^2=1+\frac{b^2}{a^2}=\frac32$ (consistent with $\frac{36}{a^2}-\frac9{b^2}=1$ giving $a^2=18$, $b^2=9$).