Let S be the focus of the hyperbola $\frac{\mathrm{x}^{2}}{3}-\frac{\mathrm{y}^{2}}{5}=1$, on the positive x -axis. Let C be the circle with its centre at $\mathrm{A}(\sqrt{6}, \sqrt{5})$ and passing through the point S . if O is the origin and SAB is a diameter of C then the square of the area of the triangle OSB is equal to

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Hyperbola

Let S be the focus of the hyperbola

\frac{x ^{2}}{3} - \frac{y ^{2}}{5} = 1

, on the positive x -axis. Let C be the circle with its centre at

A (6, 5)

and passing through the point S . if O is the origin and SAB is a diameter of C then the square of the area of the triangle OSB is equal to

Answer: 40

Hyperbola

\frac{x ^{2}}{3} - \frac{y ^{2}}{5} = 1

a^{2} = 3, b^{2} = 5, c^{2} = a^{2} + b^{2} = 8

, so

S = (22, 0)

.
Centre

A = (6, 5)

; since

S A B

is a diameter of the circle centred at

A

A

is the midpoint of

S B

B = 2 A - S = (26 - 22, 25)

.
Area of

△ O S B

with

O

at the origin and

S

on the x-axis:

Area = \frac{1}{2} ∣ O S ∣ \cdot (height of B) = \frac{1}{2} (22) (25) = 210

(Area)^{2} = 4 \times 10 = 40

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Integer answer