If for θ in [-π/3, 0 ] , the points (x, y)= (3 tan (θ+π/3 ), 2 tan (θ+π/6 ) ) lie on x y+a x+β y+γ =0 , then α 2 +β 2 +γ 2 is equal to

Mathematics · Trigonometry Ratios and Identities

JEE Main 2025 — 7 April, Morning Shift — Question 41

If for $\theta \in\left[-\frac{\pi}{3}, 0\right]$ , the points $(x, y)=$ $\left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)$ lie on $x y+a x+\beta y+\gamma$ $=0$ , then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is equal to

Option A:
80
Option B:
96
Option C:
72
Option D:
75
Correct

Answer: D

Step-by-step solution

y = 2\tan\phi \quad \mathrm{where} \quad \phi = \theta + \tfrac{\pi}{6}

\tan\phi = \tfrac{y}{2}

x = 3\tan\left(\theta + \tfrac{\pi}{3}\right) = 3\tan\left(\phi + \tfrac{\pi}{6}\right) = 3\left(\frac{\tan\phi + \tfrac{1}{\sqrt{3}}}{1 - \tan\phi \cdot \tfrac{1}{\sqrt{3}}}\right) = \frac{3\left(\tfrac{y}{2} + \tfrac{1}{\sqrt{3}}\right)}{1 - \tfrac{y}{2\sqrt{3}}}

x = \frac{(3y + 2\sqrt{3})\sqrt{3}}{2\sqrt{3} - y}

\mathrm{Given:}\; xy + \alpha x + \beta y + \gamma = 0

\frac{3(\sqrt{3}y+2)}{2\sqrt{3}-y} \cdot y \;+\; \alpha \cdot \frac{3(\sqrt{3}y+2)}{2\sqrt{3}-y} \;+\; \beta y + \gamma = 0

3(\sqrt{3}y+2)y + \alpha \cdot 3(\sqrt{3}y+2) + (\beta y + \gamma)(2\sqrt{3}-y) = 0

(3\sqrt{3}-\beta)y^2 + (6+3\sqrt{3}\alpha+2\sqrt{3}\beta-\gamma)y + (6\alpha+\gamma 2\sqrt{3}) = 0

3\sqrt{3}-\beta = 0 \quad \Rightarrow \quad \beta = 3\sqrt{3}

6\alpha + \gamma\cdot 2\sqrt{3} = 0

3\sqrt{3}\alpha - \gamma = -24

\alpha = -2\sqrt{3}, \quad \gamma = 6

\therefore \; \alpha^2 + \beta^2 + \gamma^2 = 27 + 12 + 36 = 75

\boxed{75}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Mathematics
Chapter: Trigonometry Ratios and Identities
Topic: Trigonometric Ratios of Compound Angles