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) $
$\Rightarrow b ^{2}=2 a ^{2} \ldots .(2)$
So, $e ^{2}=1+\frac{ b ^{2}}{ a ^{2}}=3$
Also, $a^{2}=\frac{9}{2} ($ from $( i ) \&( ii ))$
Thus, $\left(a^{2}, e^{2}\right)=\left(\frac{9}{2}, 3\right)$