Solve this following


Let $\mathrm{P}(3,3)$ be a point on the hyperbola,

$\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$. If the normal to it at $\mathrm{P}$ intesects

the $\mathrm{x}$-axis at $(9,0)$ and $\mathrm{e}$ is its eccentricity, then the ordered pair $\left(a^{2}, e^{2}\right)$ is equal to :

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

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

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

  4. $(9,3)$

Correct Option: 1


Since, $(3,3)$ lies on $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

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

Now, normal at $(3,3)$ is $y-3=-\frac{a^{2}}{b^{2}}(x-3)$,

which passes through $(9,0) \Rightarrow \mathrm{b}^{2}=2 \mathrm{a}^{2}$ .......(2)

So, $\mathrm{e}^{2}=1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=3$

Also, $a^{2}=\frac{9}{2}$ (from (i) & (ii))

Thus, $\left(a^{2}, e^{2}\right)=\left(\frac{9}{2}, 3\right)$


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