JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Ellipse

At the end of the latus rectum $(ae,\frac{b^2}{a})$ the normal to the ellipse is $\frac{ax}{e}-ay=a^2-b^2$. Requiring it to pass through the end of the minor axis $(0,-b)$ gives $ab=a^2-b^2$. Writing $b^2=a^2(1-e^2)$ turns this into $a^2\cdot a^2(1-e^2)=(a^2-b^2{)}^2=a^4{e}^4$, i.e. $1-e^2=e^4$, so the eccentricity satisfies $e^4+e^2-1=0$.