JEE Main 2025 Jan 28 Shift 2, Mathematics Q19: Introduction to Hyperbola

JEE Main2025Jan 28, Shift 2Mathematics

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 :

A
$4 x - 9 y = 12$
B
$x + 9 y = 36$
C
$9 x - 9 y = 32$
D
$6 x - 9 y = 20$

Solution

Eliminate $y^{2}$ between the two curves. From the circle, $y^{2} = 8 x - x^{2}$ ; substituting into $4 x^{2} - 9 y^{2} = 36$ gives $4 x^{2} - 9 (8 x - x^{2}) = 36 \Rightarrow 13 x^{2} - 72 x - 36 = 0$ , factoring as $(13 x + 6) (x - 6) = 0$ .

$x = - 6/13$ gives $y^{2} < 0$ , so reject. With $x = 6$ : $y^{2} = 12 \Rightarrow y = \pm 23$ . Thus $A = (6, 23)$ , $B = (6, - 23)$ and the midpoint of $A B$ is $(6, 0)$ .

Let $P = (α, (2 α + 4) /3)$ lie on $2 x - 3 y + 4 = 0$ . The centroid $(h, k)$ satisfies $3 h = α + 12$ and $3 k = (2 α + 4) /3$ , hence $9 k = 2 α + 4$ .

From the first relation $α = 3 h - 12$ . Substituting: $9 k = 2 (3 h - 12) + 4 = 6 h - 20$ , i.e. $6 h - 9 k = 20$ , or $6 x - 9 y = 20$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q19: Introduction to Hyperbola

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