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JEE Main 202621 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Ellipse

If the line α x+4 y= 7 , where α ∈ R , touches the ellipse 3 x^ 2...

If the line

\alpha x+4 y=\sqrt{7}

, where

\alpha \in \mathbf{R}

, touches the ellipse

3 x^{2}+4 y^{2}=1

at the point P in the first quadrant, then one of the focal distances of

P

is :

Options

A $\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{7}}$
B $\frac{1}{\sqrt{3}}-\frac{1}{2 \sqrt{5}}$
C $\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{5}}$
D $\frac{1}{\sqrt{3}}-\frac{1}{2 \sqrt{11}}$

Correct answer

A. $\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{7}}$

Step-by-step solution

Ellipse

3x^2 + 4y^2 = 1

:

a^2 = \frac{1}{3}

,

b^2 = \frac{1}{4}

,

e = \frac{1}{2}

.
Tangent at

(x_0, y_0)

:

3x_0x + 4y_0y = 1

.
Comparing with

\alpha x + 4y = \sqrt{7}

:

4y_0 = \frac{4}{\sqrt{7}} \Rightarrow y_0 = \frac{1}{\sqrt{7}}

and

3x_0 = \frac{\alpha}{\sqrt{7}}

.
Substituting in ellipse equation:

\frac{\alpha^2}{21} + \frac{4}{7} = 1 \Rightarrow \alpha^2 = 9 \Rightarrow \alpha = 3

.

P = \left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)

.
Focal distances

= a \pm ex_0 = \frac{1}{\sqrt{3}} \pm \frac{1}{2\sqrt{7}}

.
One focal distance is

\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}

.

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