Question

The sum of the squares of the perpendicular on any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ from two points on the minor axis each at a distance $\sqrt{a^2-b^2}$ from the center is $k{a}^2$, where $k=$

Answer 1

The equation of tangent to ellipse is $bxcos\theta +aysin\theta -ab=0$

The points on minor axis are $(0,\sqrt{a^2-b^2})$ and $(0,-\sqrt{a^2-b^2})$

The perpendicular distances are $\frac{a\sqrt{a^2-b^2}sin\theta -ab}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}$ and $-\frac{a\sqrt{a^2-b^2}sin\theta -ab}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}$

The sum of the squares of distances is $\frac{{2a}^2\left(\right(a^2-b^2){\sin }^2\theta +b^2)}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }=\frac{{2a}^2(b^2{\cos }^2\theta +a^2{\sin }^2\theta )}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }={2a}^2$

Therefore the value of $k$ is $2$