Question

If the ellipse $\frac{x^2}{a^2-3}+\frac{y^2}{a+4}=1$ is inscribed in a square of side length $a\sqrt{2}$ then a is

• A. 4
• B. 2
• C. 1
• D. None of these

Answer 1

The correct option is D None of these

Sides of the square will be perpendicular tangents to the ellipse so, vertices of the square will lie on director circle. So diameter of director circle is
$2\sqrt{(a^2-3)+(a+4)}=\sqrt{2a^2+2a^2}2\sqrt{a^2+a+1}=2a\Rightarrow a=-1Butforellipse{a}^2>3\&a>-4$
So there is no value of a which satisfy both.