Question

Find the range of eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$ such that the line segment joining the foci does not subtend a right angle at any point on the ellipse.

Answer 1

Compute the range of eccentricity for the given condition

Let the point on ellipse be $P(a\mathrm{cos\theta },b\mathrm{sin\theta })$

Let $A$ and $B$ be the two foci of the ellipse.

If $\angle APB=\frac{\mathrm{\pi }}{2}$, then $P$ lies on the circle having diameter $AB$.

Equation of the circle drawn on $AB$ as diameter is:

$x^2+y^2=a^2{e}^2=a^2-b^2$

$P$ should not lie on the ellipse .

$\Rightarrow$No point of intersection between $x^2+y^2=a^2-b^2$ and $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow$Radius of the circle $<$ Minor axis of parabola

$$\begin{gathered} \Rightarrow a^2-b^2<b^2 \\ \Rightarrow \frac{b^2}{a^2}>\frac{1}{2} \\ \Rightarrow 1-e^2>\frac{1}{2} \\ \Rightarrow e^2<\frac{1}{2} \\ \Rightarrow e<\frac{1}{\sqrt{2}} \end{gathered}$$

Therefore, if $e\in \left(0,\frac{1}{\sqrt{2}}\right)$, then the line segment joining the foci does not subtend a right angle at any point on the ellipse.