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For all real p, the line $\mathrm{2 p x+y \sqrt{1-p^2}=1}$ touches a fixed ellipse. Find the eccentricity of this ellipse.
Option: 1
$\frac{1}{2}$

Option: 2
$\frac{1}{\sqrt{2}}$

Option: 3
$\frac{\sqrt{3}}{2}$

Option: 4
$\sqrt{\frac{2}{3}}$

Answers (1)

Let the ellipse be $\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. }$ The line $\mathrm{y=m x \pm \sqrt{a^2 m^2+b^2}}$ touches this ellipse for all m.

Hence it is identical with $\mathrm{y=-\frac{2 p x}{\sqrt{1-p^2}}+\frac{1}{\sqrt{1-p^2}}.}$

$\mathrm{\text { Hence } m=-\frac{2 p}{\sqrt{1-p^2}} \text { and } a^2 m^2+b^2=\frac{1}{1-p^2}}$

$\mathrm{\Rightarrow a^2 \cdot \frac{4 p^2}{1-p^2}+b^2=\frac{1}{1-p^2} \quad \Rightarrow p^2\left(4 a^2-b^2\right)+b^2-1=0 \text {. }}$

This equation is true for all real p if $\mathrm{b^2=1}$ and $\mathrm{4 a^2=b^2}$ .

which is true for all p $\mathrm{\Rightarrow \mathrm{b}^2=1}$ and $\mathrm{4 \mathrm{a}^2=\mathrm{b}^2=1}$

$\mathrm{\Rightarrow b^2=1}$ and $\mathrm{a^2=1 / 4}$

Hence the ellipse is $\mathrm{\frac{x^2}{1 / 4}+\frac{y^2}{1}=1}$

If e is its eccentricity, then $\mathrm{\frac{1}{4}=1-e^2}$

$\mathrm{\Rightarrow e^2=\frac{3}{4} \quad \Rightarrow e=\frac{\sqrt{3}}{2}}$

Posted by

Ramraj Saini

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