Coordinate Geometry – Ellipse – JEE Main 21 Jan 2026 Shift 2

Solution:

Equation of ellipse: $\frac{x^2}{1/3} + \frac{y^2}{1/4} = 1$.

Here $a^2 = \frac{1}{3}, b^2 = \frac{1}{4}$.

Equation of tangent: $y = -\frac{\alpha}{4}x + \frac{\sqrt{7}}{4}$.

Condition for tangency $c^2 = a^2m^2 + b^2$:
$$\frac{7}{16} = \frac{1}{3}\left(\frac{\alpha^2}{16}\right) + \frac{1}{4}$$

$$7 = \frac{\alpha^2}{3} + 4 \Rightarrow \frac{\alpha^2}{3} = 3 \Rightarrow \alpha^2 = 9 \Rightarrow \alpha = 3$$
(Since P is in 1st quadrant, slope is negative $\Rightarrow \alpha > 0$).

Tangent: $3x + 4y = \sqrt{7}$.

Let $P(x_1, y_1)$. Tangent at P: $3xx_1 + 4yy_1 = 1$.

Comparing with $\frac{3}{\sqrt{7}}x + \frac{4}{\sqrt{7}}y = 1$:
$$x_1 = \frac{1}{\sqrt{7}}, \quad y_1 = \frac{1}{\sqrt{7}}$$

Eccentricity $e = \sqrt{1 – \frac{b^2}{a^2}} = \sqrt{1 – \frac{3}{4}} = \frac{1}{2}$.

Focal distances are $a \pm ex_1$.
$$d = \frac{1}{\sqrt{3}} \pm \frac{1}{2}\left(\frac{1}{\sqrt{7}}\right)$$

One of the distances is $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$.

Ans. (4)