Question

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

• A. $(9,3)$
• B. $(\frac{9}{2},2)$
• C. $(\frac{9}{2},3)$
• D. $(\frac{3}{2},2)$

Answer 1

The correct option is C $(\frac{9}{2},3)$

Given: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
At point $P(3,3),$ we have
$\frac{9}{a^2}-\frac{9}{b^2}=1$$\cdots \left(1\right)$
Now, normal at $P(3,3)$ is
$y-3=-\frac{a^2}{b^2}(x-3)$
Which passes through $(9,0)$
$\therefore b^2=2a^2\cdots \left(2\right)$
So, $e^2=1+\frac{b^2}{a^2}=3$
From $\left(1\right)$ and $\left(2\right),$ we have
$a^2=\frac{9}{2}$
$\therefore (a^2,e^2)=(\frac{9}{2},3)$