JEE Main 2025 Jan 23 Shift 2, Mathematics Q23: Involving Two or More Circles

JEE Main2025Jan 23, Shift 2Mathematics

The focus of the parabola $y^{2} = 4 x + 16$ is the centre of the circle $C$ of radius $5$ . If the values of $λ$ , for which $C$ passes through the point of intersection of the lines $3 x - y = 0$ and $x + λ y = 4$ , are $λ_{1}$ and $λ_{2}$ , $λ_{1} < λ_{2}$ , then $12 λ_{1} + 29 λ_{2}$ is equal to ________.

Your answer

Solution

Rewrite the parabola: $y^{2} = 4 (x + 4)$ , which is a horizontal parabola with vertex $(- 4, 0)$ . The focal length is $1$ , so the focus is at $(- 4 + 1, 0) = (- 3, 0)$ .

Circle: $(x + 3)^{2} + y^{2} = 25$ .

Find the intersection of $3 x - y = 0$ (so $y = 3 x$ ) and $x + λ y = 4$ : $x + 3 λ x = 4 \Rightarrow x = \frac{4}{1 + 3 λ}, y = \frac{12}{1 + 3 λ}$ .

Insert into the circle equation. Let $u = \frac{1}{1 + 3 λ}$ , so $x = 4 u$ , $y = 12 u$ :

$(4 u + 3)^{2} + (12 u)^{2} = 25 \Rightarrow 16 u^{2} + 24 u + 9 + 144 u^{2} = 25 \Rightarrow 160 u^{2} + 24 u - 16 = 0 \Rightarrow 20 u^{2} + 3 u - 2 = 0.$

Solving: $u = \frac{- 3 \pm 9 + 160}{40} = \frac{- 3 \pm 13}{40}$ , giving $u = \frac{1}{4}$ or $u = - \frac{2}{5}$ .

$u = \frac{1}{4}$ : $1 + 3 λ = 4 \Rightarrow λ = 1$ .

$u = - \frac{2}{5}$ : $1 + 3 λ = - \frac{5}{2} \Rightarrow λ = - \frac{7}{6}$ .

Since $λ_{1} < λ_{2}$ : $λ_{1} = - \frac{7}{6}, λ_{2} = 1$ .

$12 λ_{1} + 29 λ_{2} = 12 \cdot (- \frac{7}{6}) + 29 \cdot 1 = - 14 + 29 = 15.$

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JEE Main 2025 Jan 23 Shift 2, Mathematics Q23: Involving Two or More Circles

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