The ellipse$\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2-3}+\frac{{\mathrm{y}}^2}{\mathrm{a}+4}=1$is inscribed in a square of side length$\mathrm{a}\sqrt{2}$then they cannot take value

$Given ellipse \frac{x^2}{a^2-3}+\frac{y^2}{a+4}=1$

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

$$\begin{gathered} 2\sqrt{\left({\mathrm{a}}^2-3\right)+(\mathrm{a}+4)}=\sqrt{2{\mathrm{a}}^2+2{\mathrm{a}}^2} \\ \Rightarrow 2\sqrt{{\mathrm{a}}^2+\mathrm{a}+1}=2\mathrm{a} \\ \Rightarrow \sqrt{{\mathrm{a}}^2+\mathrm{a}+1}=\mathrm{a} \end{gathered}$$
$$\begin{gathered} \Rightarrow \mathrm{a}=-1\text{ but }{\mathrm{a}}^2\ge 3,\mathrm{a}>-4 \\ \Rightarrow \text{ a can not take }-1 \end{gathered}$$