Let P (10, 2 sqrt{15}) be a point on the hype | vedclass

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(D) $P (10, 2 \sqrt{15})$ lies on $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. $\frac{100}{a^2} - \frac{60}{b^2} = 1 \dots (1)$. Given length of latus rec

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$2700$

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(D) $P (10, 2 \sqrt{15})$ lies on $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. $\frac{100}{a^2} - \frac{60}{b^2} = 1 \dots (1)$. Given length of latus rectum $= 8$, so $\frac{2b^2}{a} = 8 \Rightarrow b^2 = 4a \dots (2)$. Substituting $(2)$ in $(1)$: $\frac{100}{a^2} - \frac{60}{4a} = 1 \Rightarrow \frac{100}{a^2} - \frac{15}{a} = 1$. $100 - 15a = a^2 \Rightarrow a^2 + 15a - 100 = 0$. $(a + 20)(a - 5) = 0$. Since $a > 0$, $a = 5$. Then $b^2 = 4(5) = 20$. The hyperbola is $\frac{x^2}{25} - \frac{y^2}{20} = 1$. Here $a^2 = 25, b^2 = 20$. Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{20}{25}} = \sqrt{\frac{45}{25}} = \frac{3\sqrt{5}}{5}$. Distance between foci $SS' = 2ae = 2(5)(\frac{3\sqrt{5}}{5}) = 6\sqrt{5}$. Area of $\Delta PSS' = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (SS') 	imes |y_P| = \frac{1}{2} 	imes 6\sqrt{5} 	imes 2\sqrt{15} = 6\sqrt{75} = 6(5\sqrt{3}) = 30\sqrt{3}$. Square of the area $= (30\sqrt{3})^2 = 900 	imes 3 = 2700$.

Let $P (10, 2 \sqrt{15})$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ whose foci are $S$ and $S'$. If the length of its latus rectum is $8$, then the square of the area of $\Delta PSS'$ is equal to:

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