Let P(3,3) be a point on the hyperbola,

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 :

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

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

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

  4. (4) $(9,3)$


Correct Option: 1

Solution:

$\because$ The equation of hyperbola is

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

$\because$ Equation of hyperbola passes through $(3,3)$

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

Equation of normal at point $(3,3)$ is :

$\frac{x-3}{\frac{1}{a^{2}} \cdot 3}=\frac{y-3}{-\frac{1}{b^{2}} \cdot 3}$

$\because$ It passes through $(9,0)$

$\frac{6}{\frac{1}{a^{2}}}=\frac{-3}{-\frac{1}{b^{2}}}$

$\therefore \frac{1}{b^{2}}=\frac{1}{2 a^{2}}$ ...(ii)

From equations (i) and (ii),

$a^{2}=\frac{9}{2}, b^{2}=9$

$\because$ Eccentricity $=e$, then $e^{2}=1+\frac{b^{2}}{a^{2}}=3$

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

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