JEE Mathematics 2025 Question with Solution

Given:

• $x = 3 \tan\left( \theta + \frac{\pi}{3} \right)$
• $y = 2 \tan\left( \theta + \frac{\pi}{6} \right)$
• The point lies on $xy + \alpha x + \beta y + \gamma = 0$

Find: $\alpha^2 + \beta^2 + \gamma^2$

Let

$$A = \theta + \frac{\pi}{3}, \qquad B = \theta + \frac{\pi}{6}$$

Then

$$A - B = \frac{\pi}{6}$$

Using

$$\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$

and substituting $\tan A = \frac{x}{3}$ and $\tan B = \frac{y}{2}$,

$$\tan\frac{\pi}{6} = \frac{\frac{x}{3} - \frac{y}{2}}{1 + \frac{x}{3}\cdot\frac{y}{2}}$$

Since

$$\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$

we get

$$\frac{1}{\sqrt{3}} = \frac{\frac{2x-3y}{6}}{\frac{6+xy}{6}} = \frac{2x-3y}{6+xy}$$

So,

$$\frac{6+xy}{\sqrt{3}} = 2x - 3y$$

Multiplying by $\sqrt{3}$,

$$6 + xy = 2\sqrt{3}x - 3\sqrt{3}y$$

Rearranging,

$$xy - 2\sqrt{3}x + 3\sqrt{3}y + 6 = 0$$

Comparing with

$$xy + \alpha x + \beta y + \gamma = 0$$

we obtain

$$\alpha = -2\sqrt{3}, \qquad \beta = 3\sqrt{3}, \qquad \gamma = 6$$

Hence,

$$\alpha^2 = 12, \qquad \beta^2 = 27, \qquad \gamma^2 = 36$$

Therefore,

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

The correct option is D.

Given:

• $x = 3\left(\frac{\tan\theta + \sqrt{3}}{1 - \sqrt{3}\tan\theta}\right)$
• $y = 2\left(\frac{\tan\theta + \frac{1}{\sqrt{3}}}{1 - \frac{\tan\theta}{\sqrt{3}}}\right)$

Find: $\alpha^2 + \beta^2 + \gamma^2$

From

$$x = 3\left(\frac{\tan\theta + \sqrt{3}}{1 - \sqrt{3}\tan\theta}\right)$$

we get

$$x - \sqrt{3}x\tan\theta = 3\tan\theta + 3\sqrt{3}$$

Hence,

$$\tan\theta = \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x}$$

Now from

$$2\left(\frac{\tan\theta + \frac{1}{\sqrt{3}}}{1 - \frac{\tan\theta}{\sqrt{3}}}\right) = y$$

we have

$$2(\sqrt{3}\tan\theta + 1) = y(\sqrt{3} - \tan\theta)$$

Substituting

$$\tan\theta = \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x}$$

into this relation and simplifying,

$$4\sqrt{3}x - 12 = y(2x + 6\sqrt{3})$$

which gives

$$xy - 2\sqrt{3}x + 3\sqrt{3}y - 6 = 0$$

Comparing with

$$xy + \alpha x + \beta y + \gamma = 0$$

we get

$$\alpha = -2\sqrt{3}, \qquad \beta = 3\sqrt{3}, \qquad \gamma = -6$$

Therefore,

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

So the correct option is D.

Note: The two extracted solution approaches differ in the sign of the constant term, but both give the same required value of $\alpha^2 + \beta^2 + \gamma^2 = 75$.