JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Straight Lines

If A and B are the points of intersection of the circle

x^{2} + y^{2} - 8 x = 0

and the hyperbola

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

and a point

P

moves on the line

2 x - 3 y + 4 = 0

, then the centroid of

△ P A B

lies on the line :

Answer: (D)

Solution

The circle is

x^{2} + y^{2} - 8 x = 0

, i.e.

y^{2} = 8 x - x^{2}

. Substituting into

4 x^{2} - 9 y^{2} = 36

(the hyperbola cleared of fractions):

4 x^{2} - 9 (8 x - x^{2}) = 36 ⟹ 13 x^{2} - 72 x - 36 = 0 ⟹ (13 x + 6) (x - 6) = 0

Only

x = 6

gives real

y

y^{2} = 48 - 36 = 12

, so

A = (6, 23)

B = (6, - 23)

.

Let

P = (α, β)

2 x - 3 y + 4 = 0

, so

β = \frac{2 α + 4}{3}

. The centroid

(h, k)

△ P A B

satisfies

h = \frac{α + 12}{3} ⟹ α = 3 h - 12, k = \frac{β}{3} ⟹ β = 3 k

Substituting into the line equation for

P

2 (3 h - 12) - 3 (3 k) + 4 = 0 ⟹ 6 h - 9 k - 20 = 0

The centroid lies on

6 x - 9 y = 20

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option