Question

If two points are taken on minor axis of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the same distance from the center as the foci, the sum of the squares of the perpendiculars from these points on any tangents to the ellipse, if $a<b$ is

• A. $a^2$
• B. $b^2$
• C. $2a^2$
• D. $2b^2$

Answer 1

Answer: (C)

$2a^2$

Given, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\therefore OS=ae=a\sqrt{1-\frac{b^2}{a^2}}=\sqrt{a^2-b^2}$

So, two points on the minor axis are $S_1(0,\sqrt{a^2-b^2}),S_2(0,-\sqrt{a^2-b^2}).$

Let tangent to the ellipse by $y=mx+c=mx+\sqrt{a^2{m}^2+b^2}$ where $m$ is parameter.

Now, sum of the squares of ${\perp }^{\prime }$s on this tangent from the points $S_1$ and $S_1^{\prime }$ is

${\left(\frac{\sqrt{a^2-b^2}-\sqrt{a^2{m}^2+b^2}}{\sqrt{1+m^2}}\right)}^2+{(-\frac{\sqrt{a^2-b^2}-\sqrt{a^2{m}^2+b^2}}{\sqrt{1+m^2}})}^2$

$=2\left(\frac{a^2-b^2+a^2{m}^2+b^2}{1+m^2}\right)=\frac{2a^2(1+m^2)}{1+m^2}=2a^2$