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(C) The sum of focal distances of a point $P(x, y)$ on a hyperbola is $2ex = 8\sqrt{\frac{5}{3}}$.Given $x = 4$,we have $2e(4) = 8\sqrt{\frac{5}{3}}$,

Vedclass · Accepted Answer

$185$

Vedclass · Answer

(C) The sum of focal distances of a point $P(x, y)$ on a hyperbola is $2ex = 8\sqrt{\frac{5}{3}}$.Given $x = 4$,we have $2e(4) = 8\sqrt{\frac{5}{3}}$,which implies $e = \sqrt{\frac{5}{3}}$.Since $e^2 = 1 + \frac{b^2}{a^2}$,we have $\frac{5}{3} = 1 + \frac{b^2}{a^2}$ $\Rightarrow \frac{b^2}{a^2} = \frac{2}{3}$ $\Rightarrow b^2 = \frac{2}{3}a^2$.Substituting $P(4, 3)$ into the hyperbola equation: $\frac{16}{a^2} - \frac{9}{b^2} = 1$.Substituting $b^2 = \frac{2}{3}a^2$: $\frac{16}{a^2} - \frac{9}{(2/3)a^2} = 1$ $\Rightarrow \frac{16}{a^2} - \frac{27}{2a^2} = 1$ $\Rightarrow \frac{32 - 27}{2a^2} = 1$ $\Rightarrow 2a^2 = 5$ $\Rightarrow a^2 = \frac{5}{2}$.Then $b^2 = \frac{2}{3} 	imes \frac{5}{2} = \frac{5}{3}$.The length of the latus rectum $l = \frac{2b^2}{a} = \frac{2(5/3)}{\sqrt{5/2}} = \frac{10}{3} 	imes \sqrt{\frac{2}{5}} = \frac{2\sqrt{10}}{3}$.$l^2 = \frac{4 	imes 10}{9} = \frac{40}{9} \Rightarrow 9l^2 = 40$.The product of focal distances $m = e^2x^2 - a^2 = \frac{5}{3}(16) - \frac{5}{2} = \frac{80}{3} - \frac{5}{2} = \frac{160 - 15}{6} = \frac{145}{6}$.Thus,$6m = 145$.Finally,$9l^2 + 6m = 40 + 145 = 185$.

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 $9l^2 + 6m$ is equal to :-

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