Ellipse
Tangent to Ellipse — Focal Distance
nta_pyq_2026_jan
Grade 11
Question:
If the line $\alpha x+4y=\sqrt{7}$, where $\alpha\in\mathbb{R}$, 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:
$\dfrac{1}{\sqrt{3}}+\dfrac{1}{2\sqrt{7}}$
$\dfrac{1}{\sqrt{3}}-\dfrac{1}{2\sqrt{5}}$
$\dfrac{1}{\sqrt{3}}+\dfrac{1}{2\sqrt{5}}$
$\dfrac{1}{\sqrt{3}}-\dfrac{1}{2\sqrt{11}}$
Step-by-Step Solution
Key Concept: Ellipse $3x^2+4y^2=1$: $a^2=\tfrac{1}{3}$, $b^2=\tfrac{1}{4}$, $e=\tfrac{1}{2}$. Tangent at $(x_0,y_0)$: $3x_0x+4y_0y=1$. Compare with $\alpha x+4y=\sqrt{7}$: $4y_0=\tfrac{4}{\sqrt{7}}$ and $3x_0=\tfrac{\alpha}{\sqrt{7}}$.
$P=\left(\tfrac{1}{\sqrt{7}},\tfrac{1}{\sqrt{7}}\right)$. Focal distance $=\tfrac{1}{\sqrt{3}}+\tfrac{1}{2\sqrt{7}}$.
Correct Answer: 1