JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Ellipse

The ellipse is $\frac{x^2}{1/3}+\frac{y^2}{1/4}=1$, so $a^2=\frac{1}{3}$, $b^2=\frac{1}{4}$. The line $\alpha x+4y=\sqrt{7}$ touches the ellipse when $c^2=a^2{m}^2+b^2$, where $m=-\frac{\alpha }{4}$. Thus $\frac{7}{16}=\frac{1}{3}\cdot \frac{{\alpha }^2}{16}+\frac{1}{4}$, giving $\frac{7}{16}-\frac{4}{16}=\frac{{\alpha }^2}{48}$, so ${\alpha }^2=9$ and $\alpha =3$ (first quadrant). The tangent at $(x_1,y_1)$ is $3x{x}_1+4y{y}_1=1$. Comparing with $3x+4y=\sqrt{7}$: $\frac{3x_1}{3}=\frac{4y_1}{4}=\frac{1}{\sqrt{7}}$, so $P(\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}})$. The eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$. The focal distance is $PS=e(\frac{a}{e}-x_1)=\frac{1}{2}(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{7}})=\frac{1}{\sqrt{3}}-\frac{1}{2\sqrt{7}}$. The other focal distance is $P{S}^{\prime }=\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}}$.