Let P ( 10 , ? 2 15 ) be a point on the hyperbola x 2 a 2 - 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 P S S ' is equal to: [2026]

Question

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 $∆ P S S'$ is equal to: [2026]

Smart Achievers · Accepted Answer

(4)

$P (10, 2 \sqrt{15}) lies on \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$∆ P S S'$ $\frac{100}{a^{2}} - \frac{60}{b^{2}} = 1 . . . (1)$

$∵$ $length of latus rectum = 8$

$\frac{2 b^{2}}{a} = 8 \Rightarrow \frac{b^{2}}{a} = 4 . . . (2)$

$From (1) & (2)$

$\frac{100}{a^{2}} - \frac{60}{4 a} = 1$

$400 - 60 a = 4 a^{2}$

$4 a^{2} + 60 a - 400 = 0$

$a^{2} + 15 a - 100 = 0$

$a = 5 & - 20 (rejected)$

$\Rightarrow b = \sqrt{20}$

$∴$ $Hyperbola is \frac{x^{2}}{25} - \frac{y^{2}}{20} = 1$

$∴$ $Focal length S_{1} S_{2} = 2 a e$ $= 2 \cdot 5 (\sqrt{1 + \frac{4}{5}}) = 6 \sqrt{5}$

$∴$ $Area of ∆ P S_{1} S_{2} = \frac{1}{2} \cdot 6 \sqrt{5} \cdot 2 \sqrt{15} = 30 \sqrt{3} = A$

$∴$ $A^{2} = 2700$

Solution

Q.
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 $∆ P S S'$ is equal to: [2026]

1 4200

2 900

3 1462

4 2700

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