Question

Let $S$ and $S^{\prime }$ be two foci of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. If a circle described on $S{S}^{\prime }$ as diameter intersects the ellipse in real and distinct points, then the eccentricity $e$ of the ellipse satisfies

• A. $e=\frac{1}{\sqrt{2}}$
• B. $e\in (\frac{1}{\sqrt{2}},1)$
• C. $e\in (0,\frac{1}{\sqrt{2}})$
• D. None of these

Answer 1

Answer: (B)

$e\in (\frac{1}{\sqrt{2}},1)$

Given equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The foci of the ellipse are $S(ae,0)$ and $S^{\prime }(-ae,0)$

Equation of circle with $S{S}^{\prime }$ as diameter is

$(x+ae)(x-ae)+(y-0)(y-0)=0x^2-a^2{e}^2+y^2=0x^2+y^2=a^2{e}^2$

$\Rightarrow y^2=a^2{e}^2-x^2$

Substituting in the equation of ellipse, we have

$\frac{x^2}{a^2}+\frac{a^2{e}^2-x^2}{b^2}=1\Rightarrow x^2(\frac{1}{a^2}-\frac{1}{b^2})=1-\frac{a^2{e}^2}{b^2}\Rightarrow x^2\left(\frac{b^2-a^2}{b^2{a}^2}\right)=\frac{b^2-a^2{e}^2}{b^2}\Rightarrow x^2\left(\frac{b^2-a^2}{a^2}\right)=b^2-a^2{e}^2\Rightarrow x^2(\frac{b^2}{a^2}-1)=a^2-a^2{e}^2-a^2{e}^2{x}^2(-e^2)=a^2(1-2e^2)\Rightarrow x^2=\frac{a^2}{e^2}(2e^2-1)\Rightarrow x=\pm \frac{a}{e}\sqrt{2e^2-1}$

Circle intersect the ellipse on distinct points so the values of $x$ will be real and distinct, if

$2e^2-1>0e^2>\frac{1}{2}e>\pm \frac{1}{\sqrt{2}}$

But $e$ always lies between $(0,1)$.

$\therefore e\in (\frac{1}{\sqrt{2}},1)$