Aspire Faculty ID #17913 · Topic: JEE Main 2026 (21 January Evening Shift) · Just now
JEE Main 2026 (21 January Evening Shift)

In the line $ax + 4y = \sqrt{7}$, where $a \in R$, touches the ellipse $3x^2 + 4y^2 = 1$ at the point $P$ in the first quadrant, then one of the focal distances of $P$ is:

Solution

$ax + 4y - \sqrt{7} = 0$ touches $3x^2 + 4y^2 = 1$ 
$\Rightarrow \frac{c^2}{a^2} = \frac{7}{16} = \frac{1}{3}\cdot \frac{a^2}{16} + \frac{1}{4}$ 
$\Rightarrow a = 3, -3$ 
Tangent is $3x + 4y - \sqrt{7} = 0$ 
Let the point of contact is $P(x_1, y_1)$ 
$\Rightarrow 3x_1 + 4y_1 = \sqrt{7}$ 
$\Rightarrow \frac{3x_1}{3} + \frac{4y_1}{4} = \frac{\sqrt{7}}{1}$ 
$\Rightarrow P\left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)$ 
$e = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$ 
$PS = e \cdot PM$ 
$\Rightarrow PS = \frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$

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