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 Δ PSS^ is equal to :

Correct answer: D. 2700 For the hyperbola (x^2)/(a^2) - (y^2)/(b^2) = 1 , point P(10, 2 15 ) gives (100)/(a^2) - (60)/(b^2) = 1 . The latus rectum length (2b^2)/(a) = 8 gives b^2 = 4a . Substituting: (100)/(a^2) - (15)/(a) = 1 , which yields a^2 + 15a - 100 = 0 , so a = 5 and b^2 = 20 . Thus c^2 = 45 and c = 3 5 . The foci are S(± 3 5 , 0) . Triangle PSS' has base SS' = 6 5 and height 2 15 . Area = (1)/(2) · 6 5 · 2 15 = 30 3 . Therefore (Area) ^2 = 2700 .

JEE Main 202622 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Hyperbola

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Hyperbola

Let P(10,2 15 ) be a point on the hyperbola x^ 2 a^ 2 - y^ 2 ~b^ 2...

Let

\mathrm{P}(10,2 \sqrt{15})

be a point on the hyperbola

\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1

, whose foci are S and

\mathrm{S}^{\prime}

. If the length of its latus rectum is 8, then the square of the area of

\Delta \mathrm{PSS}^{\prime}

is equal to :

Options

A900
B4200
C1462
D2700

Correct answer

D. 2700

Step-by-step solution

For the hyperbola

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

, point

P(10, 2\sqrt{15})

gives

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

.
The latus rectum length

\frac{2b^2}{a} = 8

gives

b^2 = 4a

.
Substituting:

\frac{100}{a^2} - \frac{15}{a} = 1

, which yields

a^2 + 15a - 100 = 0

, so

a = 5

and

b^2 = 20

.
Thus

c^2 = 45

and

c = 3\sqrt{5}

.
The foci are

S(\pm 3\sqrt{5}, 0)

.
Triangle

PSS'

has base

SS' = 6\sqrt{5}

and height

2\sqrt{15}

.
Area

= \frac{1}{2} \cdot 6\sqrt{5} \cdot 2\sqrt{15} = 30\sqrt{3}

.
Therefore (Area)

^2 = 2700

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