Coordinate Geometry – Conic Sections – JEE Main 28 Jan 2025 Shift 2

Solution:

First, find points A and B.
Circle: $x^2 + y^2 – 8x = 0 \Rightarrow y^2 = 8x – x^2$.
Hyperbola: $4x^2 – 9y^2 = 36$.
Substitute $y^2$:
$$4x^2 – 9(8x – x^2) = 36$$
$$4x^2 – 72x + 9x^2 = 36$$
$$13x^2 – 72x – 36 = 0$$
$$(13x + 6)(x – 6) = 0$$
$x = 6$ or $x = -6/13$.
If $x = -6/13$, $y^2 = 8(-6/13) – (-6/13)^2 < 0$, so reject. So, $x = 6$.
Substitute $x=6$ into circle equation:
$$36 + y^2 – 48 = 0 \Rightarrow y^2 = 12 \Rightarrow y = \pm\sqrt{12} = \pm 2\sqrt{3}$$
Points are $A(6, 2\sqrt{3})$ and $B(6, -2\sqrt{3})$.

Let point P on the line $2x – 3y + 4 = 0$ be $P(\alpha, \beta)$.
Then $2\alpha – 3\beta + 4 = 0 \Rightarrow \beta = \frac{2\alpha+4}{3}$.
So $P(\alpha, \frac{2\alpha+4}{3})$.

Let centroid of $\Delta PAB$ be $G(h, k)$.
$$h = \frac{x_A + x_B + x_P}{3} = \frac{6 + 6 + \alpha}{3} = \frac{12 + \alpha}{3}$$
$$\Rightarrow \alpha = 3h – 12$$

$$k = \frac{y_A + y_B + y_P}{3} = \frac{2\sqrt{3} – 2\sqrt{3} + \beta}{3} = \frac{\beta}{3}$$
$$\Rightarrow \beta = 3k$$

Since P lies on $2x – 3y + 4 = 0$, substitute $\alpha$ and $\beta$:
$$2(3h – 12) – 3(3k) + 4 = 0$$
$$6h – 24 – 9k + 4 = 0$$
$$6h – 9k – 20 = 0$$

Replacing $(h, k)$ with $(x, y)$, the locus is:
$$6x – 9y = 20$$

Ans. (4)