Question

The circle on $S{S}^1$(distance between foci) as diameter intersects the ellipse in real points then its eccentricity is

• A. $e=\frac{1}{\sqrt{2}}$
• B. $e<\frac{1}{\sqrt{2}}$
• C. $e>\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is C $e>\frac{1}{\sqrt{2}}$

As radius $r=ae$

From diagram

$ae>b$

$e^2>\frac{b^2}{a^2}$

we know

$e^2=1-\frac{b^2}{a^2}$

So,

$1-\frac{b^2}{a^2}>\frac{b^2}{a^2}$

$2\frac{b^2}{a^2}<1$

$\frac{b^2}{a^2}<\frac{1}{2}$

Multiplying with (-1) both side

$-\frac{b^2}{a^2}>-\frac{1}{2}$ (negative sign changes less than to greater than)

Now adding 1 both side

$1-\frac{b^2}{a^2}>1-\frac{1}{2}$

$e^2>\frac{1}{2}$

$e>\frac{1}{\sqrt{2}}$