Question

If the ellipse $\frac{x^2}{a^2-7}+\frac{y^2}{13-5a}=1$ is inscribed in a square of side length $\sqrt{2}a$, then $a$ is equal to:

• A. $\frac{6}{5}$
• B. $(-\infty ,-\sqrt{7})\cup (\sqrt{7},\frac{13}{5})$
• C. $(-\infty ,-\sqrt{7})\cup (\frac{13}{5},\sqrt{7})$
• D. No such $a$ exists

Answer 1

The correct option is D No such $a$ exists

We know if perpendicular tangents are drawn on an ellipse from a point then the point lie on the director circle.

Here, $AB=2\sqrt{a^2+b^2}$

$\Rightarrow AB=2\sqrt{a^2-7+13-5a}$

$\Rightarrow AB=2\sqrt{a^2-5a+6}$

In $△ABC,$

By Pythagoras theorem we get,

${\left(AB\right)}^2={\left(AC\right)}^2+{\left(BC\right)}^2$

$\Rightarrow 4(a^2-5a+6)=2a^2+2a^2$

$\Rightarrow a=\frac{6}{5}$

But If $a=\frac{6}{5}$, then the term $(a^2-7)$ becomes negative which is not possible.

$\therefore$ No such $a$ exists.