JEE Main 2020 (Online) 4th September Morning ShiftQuestion 1
Let $P(3,3)$ be a point on the hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal to it at $P$ intersects the $x$-axis at $(9,0)$ and $e$ is its eccentricity, then the ordered pair $(a^2,e^2)$ is equal to :
Hyperbola
Solution
Tangent slope at $(3,3)$ is $\frac{b^2}{a^2}$, so the normal slope is $-\frac{a^2}{b^2}$; through $(3,3)$ and $(9,0)$ it is $-\frac12$, hence $b^2=2a^2$. Then $\frac9{a^2}-\frac9{2a^2}=1$, $a^2=\frac92$, $b^2=9$, $e^2=1+\frac{b^2}{a^2}=3$. So $(a^2,e^2)=\left(\frac92,3\right)$.