Question

The line $2px+y\sqrt{\left(1-p^2\right)}=1,\left(\left|p\right|<1\right)$ for different values of $p$, touches a fixed ellipse whose axes are the coordinate axes.
The eccentricity of the ellipse is :

• A. $\frac{1}{\sqrt{5}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{2}{\sqrt{5}}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}$

Let the ellipse be

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

The line $y=mx\pm \sqrt{a^2{m}^2+b^2}$ touches, the ellipse for all $m$.

Hence, it is identical with $y=-\frac{2px}{\sqrt{1-p^2}}+\frac{1}{\sqrt{1-p^2}}$

Hence ,$m=-\frac{2p}{\sqrt{1-p^2}}$

or $a^2{m}^2+b^2=\frac{1}{1-p^2}$

or $a^2{(-\frac{2p}{\sqrt{1-p^2}})}^2+b^2=\frac{1}{1-p^2}$

or $\frac{4p^2{a}^2}{1-p^2}+b^2=\frac{1}{1-p^2}$

or $\frac{4p^2{a}^2}{1-p^2}+b^2-\frac{1}{1-p^2}=0$

or $\frac{4p^2{a}^2-1}{1-p^2}+b^2=0$

or $4p^2{a}^2-1=-b^2(1-p^2)$

or $4p^2{a}^2-1=b^2{p}^2-b^2$

or $4p^2{a}^2-b^2{p}^2=1-b^2$

or $p^2(4a^2-b^2)+b^2-1=0$

This equation is true for all real $p$ if

$b^2=1$ and $4a^2=b^2$

$b^2=1$ and $a^2=\frac{1}{4}$

If $e$ is its eccentricity, then

$\frac{1}{4}=1-e^2$ or $e^2=1-\frac{1}{4}=\frac{4-1}{4}=\frac{3}{4}$

$\therefore$eccentricity$=e=\frac{\sqrt{3}}{2}$

Option$\left(c\right)$ is correct.