JEE Main 2025 Jan 23 Shift 2, Mathematics Q5: Points and Straight Lines

JEE Main2025Jan 23, Shift 2Mathematics

A rod of length eight units moves such that its ends $A$ and $B$ always lie on the lines $x - y + 2 = 0$ and $y + 2 = 0$ , respectively. If the locus of the point $P$ , that divides the rod $A B$ internally in the ratio $2 : 1$ is $9 (x^{2} + α y^{2} + β x y + γ x + 28 y) - 76 = 0$ , then $α - β - γ$ is equal to :

A
24
B
23
C
21
D
22

Solution

Let $A = (α, α + 2)$ on $y = x + 2$ and $B = (β, - 2)$ on $y = - 2$ . The point $P (h, k)$ divides $A B$ in $2 : 1$ internally:

$h = \frac{2 β + α}{3}, k = \frac{2 ( - 2 ) + ( α + 2 )}{3} = \frac{α - 2}{3} .$

From the $k$ equation: $α = 3 k + 2$ . From the $h$ equation: $2 β = 3 h - α = 3 h - 3 k - 2$ , so $β = \frac{3 h - 3 k - 2}{2}$ .

The rod length condition $∣ A B ∣ = 8$ gives $(α - β)^{2} + (α + 4)^{2} = 64$ . Substituting and simplifying yields

$9 (x^{2} + 13 y^{2} - 6 x y - 4 x + 28 y) = 76.$

Matching coefficients: $α = 13, β = - 6, γ = - 4$ . Therefore $α - β - γ = 13 - (- 6) - (- 4) = 23$ .

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JEE Main 2025 Jan 23 Shift 2, Mathematics Q5: Points and Straight Lines

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