Question

If two points are taken on the minor axis of an ellipse at the same distance from the center as the foci, then the sum of the squares of the perpendicular distances from these points on any tangent to the ellipse is –

• A. 2a²
• B. 3a²
• C. 4a²
• D. None of these

Answer 1

Answer: (A)

2a²

Answer 2

Let the equation of the ellipse be

Then, the distance of a focus from the centre = ae

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

So that the two points on the minor axis are

S₁(0, $\sqrt{a^{2} - b^{2}}$) and S₁¢(0, –$\sqrt{a^{2} - b^{2}}$).

Now any tangent to the ellipse is

y = mx + $\sqrt{a^{2}m^{2} + b^{2}}$, where m is aparameter.

The sum of the squares of the perpendiculars on this tangent from the two points

S₁ and S₁¢ = $\left( \frac{\sqrt{a^{2} - b^{2}} - \sqrt{a^{2}m^{2} + b^{2}}}{\sqrt{1 + m^{2}}} \right)^{2}$

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

= $\frac{2(a^{2} - b^{2} + a^{2}m^{2} + b^{2})}{1 + m^{2}}$ = 2a² (constant).

Hence (1) is the correct answer.