Question

The sum of the squares of perpendicuars on any tangents of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$ from two points on minor axis each one at a distance of $\sqrt{a^2-b^2}$ unit from the centre is

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

Answer 1

The correct option is B $2a^2$

For the given ellipse
$\sqrt{a^2-b^2}=ae\cdots \left(1\right)$,
where $e$ is eccentricity of the ellipse.
Now coordinates of the points on the minor axis which is at distance of $ae$ from the centre of the ellipse is
$(0,\pm ae)$
Let any tangent to the given ellipse is
$y=mx+\sqrt{a^2{m}^2+b^2}$
Now length of the perpendiculars from $(0,\pm ae)$ are $d_1,d_2$, then
$d_1^2+d_2^2=\frac{(ae-\sqrt{a^2{m}^2+b^2}{)}^2}{1+m^2}+\frac{(-ae-\sqrt{a^2{m}^2+b^2}{)}^2}{1+m^2}=\frac{2(a^2{e}^2+a^2{m}^2+b^2)}{1+m^2}\Rightarrow d_1^2+d_2^2=\frac{2(a^2+a^2{m}^2)}{1+m^2}=2a^2\left[\text{using}\right(1\left)\right]$