Q45 / 75 · JEE Main 2026 (21 January Shift 2) · Mathematics
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 :
- Correct answerA$\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{7}}$
- B$\frac{1}{\sqrt{3}}-\frac{1}{2 \sqrt{5}}$
- C$\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{5}}$
- D$\frac{1}{\sqrt{3}}-\frac{1}{2 \sqrt{11}}$
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 $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$.
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 $\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}$.
Hence, the correct answer is option: 'A'.