Question

Perpendiculars are drawn from the points $(0,\pm ae)$ on any tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then the sum of their squares is:

• A. $2b^2$
• B. $2a^2$
• C. $b^2$
• D. $4a^2$

Answer 1

Answer: (B)

$2a^2$

The equation of tangent of ellipse is $\frac{x\cos \varphi }{a}+\frac{y\sin \varphi }{b}=1$

Sum of squares of perpendiculars is

$\frac{(1-\frac{ae\sin \varphi }{b}{)}^2+(1+\frac{ae\sin \varphi }{b}{)}^2}{\frac{{\cos }^2\varphi }{a^2}+\frac{{\sin }^2\varphi }{b^2}}$

$=2a^2\left(\frac{b^2+a^2{e}^2{\sin }^2\varphi }{b^2{\cos }^2\varphi +a^2{\sin }^2\varphi }\right)$

$=2a^2\left(\frac{b^2+a^2{e}^2{\sin }^2\varphi }{b^2+(a^2-b^2){\sin }^2\varphi }\right)$

But $a^2{e}^2=a^2-b^2$

$\Rightarrow$ sum of squares $=2a^2$