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 $\left(a^2,e^2\right)$ is equal to :

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

Answer 1

Answer: (A)

Since, (3,3) lies on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{9}{a^2}-\frac{9}{b^2}=1\dots$(1)
Now, normal at (3,3) is $y-3=-\frac{a^2}{b^2}(x-3)$
which passes through $(9,0)$
$\Rightarrow b^2=2a^2\dots .(2)$
So, $e^2=1+\frac{b^2}{a^2}=3$
Also, $a^2=\frac{9}{2}($ from $(i)\&(ii))$
Thus, $\left(a^2,e^2\right)=\left(\frac{9}{2},3\right)$