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 centre is:

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

Answer 1

Answer: (C)

$2a^2$

The equation of tangent to ellipse is $bx\cos \theta +ay\sin \theta =ab$

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

The length of perpendicular from those two points on tangent are $\frac{a(\sqrt{a^2-b^2}\sin \theta -b)}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}$ and $\frac{a(-\sqrt{a^2-b^2}\sin \theta -b)}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}$

The sum of squares of perpendiculars is $\frac{2a^2(b^2{\cos }^2\theta +a^2{\sin }^2\theta )}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }=2a^2$

Therefore, correct option is C.