JEE Main 2025 — 8 April, Evening Shift — Question 22

$S: x^{2}+y^{2}-2 x-4 y-11=0$

Centre $C(1,2)$

radius $=\sqrt{1+4+11}$ = 4

Ellipse $3 x^{2}+p y^{2}=4$ passes through $(1,2)$

$3(1)+p(4)=4$

$4 p=1$

$p=\frac{1}{4}$

$E: 3 x^{2}+\frac{y^{2}}{4}=4$

or $\frac{x^{2}}{\frac{4}{3}}+\frac{y^{2}}{16}=1$

$e=\sqrt{1-\frac{\frac{4}{3}}{16}}$

$e=\sqrt{1-\frac{1}{12}}$

$e=\sqrt{\frac{11}{12}}$

Focus $=\left(0, \pm 2 \sqrt{\frac{11}{3}}\right)$

$f_{1}=\sqrt{1+\left(2-2 \sqrt{\frac{11}{3}}\right)^{2}}$ and $f_{2}=\sqrt{1+\left(2+2 \sqrt{\frac{11}{3}}\right)^{2}}$

$f_{1} f_{2}=\frac{37}{3}$

$\Rightarrow 6 f_{1} f_{2}-r=74-7$

$=70$