Question

If normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(ae,\frac{b^2}{a})$ is passing through $(0,-2b)$, then $e=$

• A. $\frac{1}{2}$
• B. $2(\sqrt{2}-1)$
• C. $\sqrt{2\sqrt{2}-2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $2(\sqrt{2}-1)$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{2x}{a^2}+\frac{2y}{b^2}\cdot y^{\prime }=0$

$\Rightarrow y^{\prime }=\frac{-x}{a^2}\times \frac{b^2}{y}=\frac{-x{b}^2}{y{a}^2}$ at $(ae,\frac{b^2}{a})$

$y^{\prime }=\frac{-ae{b}^2}{b^{2}a}=-e$

$\Rightarrow \frac{-1}{y}=\frac{1}{e}$

$\Rightarrow$ Slope of normal, $m=\frac{1}{e}$

$\Rightarrow$ Equation of normal $y-\frac{b^2}{a}=\frac{1}{e}(x-ae)$

$\Rightarrow (ay-b^2)e=ax-a^{2}e$

It passes $(0,-2b)$

$\Rightarrow (-2ab-b^2)e=-a^{2}e$

$\Rightarrow e=\frac{a^3}{b(2a+b)}$

$\Rightarrow 2abe+b^{2}e=-a^{2}e$

$\Rightarrow -a^2+b^2+2ab=0$

$\Rightarrow \frac{-a^2}{b^2}+1+\frac{2a}{b}=0$

$\frac{a^2}{b^2}-\frac{2a}{b}-1=0$

Let $\frac{a}{b}=t$

$\Rightarrow t^2-2t-1=0$

$\Rightarrow t=\frac{2\pm \sqrt{4+4}}{2}$

$=\frac{2\pm 2\sqrt{2}}{2}$

$=1\pm \sqrt{2}$

$=1+\sqrt{2}$

$\{\frac{a}{b}>0\}$

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

$\Rightarrow e^2=1-\frac{1}{t^2}=1-\frac{1}{(3-2\sqrt{2})}$

$=\frac{(2+2\sqrt{2})}{3+2\sqrt{2}}\frac{(3-2\sqrt{2})}{(3-2\sqrt{2})}$

$\Rightarrow e^2=\frac{6+6\sqrt{2}-4\sqrt{2}-8}{9-8}=-2+2\sqrt{2}$

$=2(\sqrt{2}-1)$

$\Rightarrow \left(B\right)$.