Question

Let $P(3,3)$ be a point on the hyperbola, $\left(\frac{x^2}{a^2}\right)-\left(\frac{y^2}{b^2}\right)=1$. 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. $\left(9,3\right)$
• B. $\left(\frac{9}{2},2\right)$
• C. $\left(\frac{9}{2},3\right)$
• D. $\left(\frac{3}{2},2\right)$

Answer 1

The correct option is C

$\left(\frac{9}{2},3\right)$

Explanation for the correct option:

Finding the values for the ordered pair:

Given point $P\left(3,3\right)$ lies on the hyperbola $\left(\frac{x^2}{a^2}\right)-\left(\frac{y^2}{b^2}\right)=1$. Therefore,

$\left(\frac{9}{a^2}\right)-\left(\frac{9}{b^2}\right)=1\left(1\right)$

Equation of normal is given by,

$\frac{a^{2}x}{3}+\frac{b^{2}y}{3}=a^2{e}^2$

This normal intersects the x-axis at $\left(9,0\right)$. Hence,

$$\begin{gathered} \frac{a^{2}x}{3}+0=a^2{e}^2 \\ \Rightarrow \frac{9}{3}=e^2 \\ \Rightarrow e^2=3 \end{gathered}$$

Also, we know that,

$$\begin{gathered} e^2=1+\frac{b^2}{a^2} \\ \Rightarrow 1+\frac{b^2}{a^2}=3 \\ \Rightarrow b^2=2a^2 \end{gathered}$$

Substituting this in equation $\left(1\right)$ we get,

$$\begin{gathered} \frac{9}{a^2}-\frac{9}{2a^2}=1 \\ \Rightarrow \frac{9}{a^2}\left(1-\frac{1}{2}\right)=1 \\ \Rightarrow \frac{9}{a^2}=2 \\ \Rightarrow a^2=\frac{9}{2} \end{gathered}$$

Therefore, the ordered pair $\left(a^2,e^2\right)$ is equal to $\left(\frac{9}{2},3\right)$

Hence, the correct answer is option (C).