The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and , respectively. Let the line touch this hyperbola at . If m is the product of the focal distances of the point , then is equal to __________. [2024]

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Solution

Q.
The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and $x = \pm \frac{4}{\sqrt{3}}$ , respectively. Let the line $y - \sqrt{3} x + \sqrt{3} = 0$ touch this hyperbola at $(x_{0}, y_{0})$ . If m is the product of the focal distances of the point $(x_{0}, y_{0})$ , then $4 e^{2} + m$ is equal to __________. [2024]

Ans.
**(*)

Equation of tangent to hyperbola is given by $y = m x \pm \sqrt{a^{2} m^{2} - b^{2}}$

$\Rightarrow m = \sqrt{3}$

and $a^{2} m^{2} - b^{2} = 3$

$\Rightarrow 3 a^{2} - b^{2} = 3$ ... (i)

Now, length of latus rectum of hyperbola = 9

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

$\Rightarrow b^{2} = \frac{9 a}{2}$ ... (ii)

From equation (i) and (ii), we get

$3 a^{2} - \frac{9 a}{2} = 3$

$\Rightarrow 6 a^{2} - 9 a - 6 = 0 \Rightarrow 2 a^{2} - 3 a - 2 = 0$

$\Rightarrow 2 a^{2} - 4 a + a - 2 = 0 \Rightarrow (2 a + 1) (a - 2) = 0$

$\Rightarrow a = 2, \frac{- 1}{2} (ignore) \Rightarrow a = 2 and b = 3$

Equation of hyperbola is $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$

Solving for tangent $y = \sqrt{3} x - \sqrt{3}$ , we get

$\frac{x^{2}}{4} - \frac{3 x^{2} + 3 - 6 x}{9} = 1$

$\Rightarrow 9 x^{2} - 12 x^{2} - 12 + 24 x = 36$

$\Rightarrow 3 x^{2} - 24 x + 48 = 0$

$\Rightarrow x^{2} - 8 x + 16 = 0 \Rightarrow {(x - 4)}^{2} = 0$

$\Rightarrow x = 4 and y = 3 \sqrt{3}, e = \sqrt{1 + \frac{9}{4}} = \frac{\sqrt{13}}{2}$

So, $(x_{0}, y_{0}) \equiv (4, 3 \sqrt{3})$ is the point of contact.

Focus of hyperbola is $(\pm \sqrt{13}, 0)$

$∴ P F_{1} \cdot P F_{2}$

$= (\sqrt{{(4 - \sqrt{13})}^{2} + {(3 \sqrt{3})}^{2}}) (\sqrt{{(4 + \sqrt{13})}^{2} + {(3 \sqrt{3})}^{2}})$

$= \sqrt{2304} = 48 = m$

$∴ 4 e^{2} + m = 4 \times 13 / 4 + 48 = 61$

Note:** Here equation of directrix should be $x = \pm \frac{4}{\sqrt{13}}$ , but in given question it is given $x = \pm \frac{4}{\sqrt{3}}$ which is wrong because eccentricity should be greater than 1, So, this question is bonus.

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