Let the sum of the focal distances of the point P(4, 3) on the hyperbola be . If for H, the length of the latus rectum is l and the product of the focal distances of the point P is m, then is equal to : [2025]

Question

Let the sum of the focal distances of the point P(4, 3) on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be $8 \sqrt{\frac{5}{3}}$ . If for H, the length of the latus rectum is $l$ and the product of the focal distances of the point P is m, then $9 l^{2} + 6 m$ is equal to : [2025]

Smart Achievers · Accepted Answer

(3)

Given, hyperbola $(H) : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

Sum of focal distance of P(4, 3) = $8 \sqrt{\frac{5}{3}}$

$\Rightarrow 2 e x = 8 \sqrt{\frac{5}{3}} \Rightarrow 2 e (4) = 8 \sqrt{\frac{5}{3}} \Rightarrow e^{2} = \frac{5}{3}$

$∵ b^{2} = a^{2} (\frac{5}{3} - 1) \Rightarrow b^{2} = \frac{2}{3} a^{2}$ ... (i)

$\Rightarrow \frac{16}{a^{2}} - \frac{9}{b^{2}} = 1$ ... (ii)

Using (i) and (ii), we get

$\Rightarrow a^{2} = \frac{5}{2}, b^{2} = \frac{5}{3}$

Now, length of latus rectum, $l = \frac{2 b^{2}}{a} \Rightarrow l^{2} = \frac{4 b^{4}}{a^{2}}$

$\Rightarrow l^{2} = \frac{4 \times {(5 / 3)}^{2}}{5 / 2} \Rightarrow 9 l^{2} = 40$

Also, $m = (e x + a) (e x - a) = {(e x)}^{2} - {(a)}^{2}$

$= \frac{5}{3} (16) - \frac{5}{2} = \frac{145}{6}$

$\Rightarrow 6 m = 145$

$∴ 9 l^{2} + 6 m = 40 + 145 = 185$ .

Solution

1 187

2 184

3 185

4 186

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