JEE Main 2026 — 21 January, Evening Shift — Question 4

$\alpha x+4 y-\sqrt{7}=0$ touches $3 x^{2}+4 y^{2}=1$

$\therefore \mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}$

$\frac{7}{16}=\frac{1}{3} \times \frac{\alpha^{2}}{16}+\frac{1}{4} \Rightarrow \alpha=3,-3$

Tangent is $3 x+4 y-\sqrt{7}=0$

Let the point of contact is $\mathrm{P}\left(\mathrm{x}_{1} \mathrm{y}_{1}\right)$

∴ Tangent is $3 \mathrm{xx}_{1}+4 \mathrm{yy}_{1}=1$

$\therefore \frac{3 \mathrm{x}_{1}}{3}=\frac{4 \mathrm{y}_{1}}{4}=\frac{1}{\sqrt{7}} \quad$

$\therefore \mathrm{P}\left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)$

$\mathrm{e}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$

$\mathrm{PS}=\mathrm{e}(\mathrm{PM})$

$\mathrm{PS} =e\left(\frac{a}{e}-\frac{1}{\sqrt{7}}\right)$

$\mathrm{PS} =\frac{1}{2}\left(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{7}}\right)$

$\mathrm{PS}=\frac{1}{\sqrt{3}}-\frac{1}{2 \sqrt{7}}$

$\mathrm{ P S}^{\prime}=e\left(P M^{\prime}\right)=\frac{1}{2}\left(\frac{a}{e}+\frac{1}{\sqrt{7}}\right)$

$\mathrm{PS}^{\prime}=\frac{1}{2}\left(\frac{1}{\sqrt{7}}+\frac{2}{\sqrt{3}}\right)$

$\mathrm{PS}^{\prime}=\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{7}}$