JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Ellipse

The first line meets the $x$-axis when $y=0$: $\frac{x}{7}=1$, so $x=7$. Thus $(7,0)$ is on the ellipse, giving $a=7$. The second line meets the $y$-axis when $x=0$: $-\frac{y}{2\sqrt{6}}=1$, so $y=-2\sqrt{6}$. Thus $(0,-2\sqrt{6})$ is on the ellipse, giving $b=2\sqrt{6}$. The eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{24}{49}}=\sqrt{\frac{25}{49}}=\frac{5}{7}$.