#### For all $\mathrm{p \in[-1,1]}$, the line $\mathrm{2 p x+y \sqrt{1-p^2}=1}$ touches a fixed ellipse. Find the eccentricity of the ellipse.  Option: 1 $\frac{1}{\sqrt{2}}$Option: 2 $\frac{1}{2}$Option: 3 $\frac{\sqrt{3}}{4}$Option: 4 $\frac{\sqrt{3}}{2}$

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 $\mathrm{ y=\frac{-2 p x}{\sqrt{1-p^2}}+\frac{1}{\sqrt{1-\rho^2}}}$

Hence $\mathrm{ m=-\frac{2 p}{\sqrt{1-p^2}}}$ and $\mathrm{ a^2 m^2+b^2=\frac{1}{1-p^2} \Rightarrow a^2 \cdot \frac{4 p^2}{1-p^2}+b^2=\frac{1}{1-p^2}}$
$\mathrm{=p^2\left(4 a^2-b^2\right)+b^2-1=0}$
This equation is true for all $\mathrm{p \in[-1,1]; if ~b^2=1~ and ~4 a^2=b^2}$.

$\mathrm{\Rightarrow \quad b^2=1 ~and ~a^2=\frac{1}{4}}$.

Hence the ellipse is $\mathrm{\frac{x^2}{1 / 4}+\frac{y^2}{1}=1}$
If e its eccentricity, then $\mathrm{\frac{1}{4}=1-e^2 \Rightarrow e^2=\frac{3}{4} \Rightarrow e-\frac{\sqrt{3}}{2}.}$