Mathematics - Ellipse Question with Solution | TestHub

The sides of the square will be perpendicular tangents to the ellipse, so the vertices of the square will lie on the director circle. The diameter of the director circle is given by:

$$2\sqrt{\left(a^2-3\right)+(a+4)}=\sqrt{2a^2+2a^2}$$

This simplifies to $2\sqrt{a^2+a+1}=2a$, which implies $a=-1$.

However, for an ellipse, the conditions are $a^2>3$ and $a>-4$. Therefore, $a$ cannot take the value $-1$.