Question

The sum of the squares of the perpendiculars 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 distances $ae$ from the centre, is

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

Answer 1

Answer: (A)

$2a^2$

Any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ having slope m is $y=mx+\sqrt{a^2{m}^2+b^2}$
Points on the minor axis are $(0,ae),(0,-ae)$
$\therefore$ Sum of the squares of the perpendicular on the tangent from $(0,ae)$ and $(0,-ae)$

$={\left|\frac{\sqrt{a^2{m}^2+b^2}-ae}{\sqrt{m^2+1}}\right|}^2+{\left|\frac{\sqrt{a^2{m}^2+b^2}+ae}{\sqrt{m^2+1}}\right|}^2$

$=\frac{2(a^2{m}^2+b^2+a^2{e}^2)}{m^2+1}$

$=\frac{2(a^2{m}^2+a^2-a^2{e}^2+a^2{e}^2)}{m^2+1}$

$=\frac{2a^2(m^2+1)}{m^2+1}=2a^2$

Hence, option A.