Question

The line $2px+y\sqrt{1-p^2}=1(|p|<1)$ for different value of p touches

• A. an ellipse of eccentricity $\frac{2}{\sqrt{3}}$
• B. an ellipse of eccentricity $\frac{\sqrt{3}}{2}$
• C. hyperbola of eccentricity 2
• D. None of the above

Answer 1

The correct option is B an ellipse of eccentricity $\frac{\sqrt{3}}{2}$

$y=\frac{-2p}{\sqrt{1-p^2}}x+\frac{1}{\sqrt{1-p^2}}$
$m=-\frac{2p}{\sqrt{1-p^2}}\Rightarrow p^2=\frac{m^2}{4+m^2}$
$y=mx+\frac{1}{\sqrt{1-\frac{m^2}{4+m^2}}}\Rightarrow y=mx+\sqrt{\frac{4+m^2}{4}}$
$\Rightarrow y=mx+\sqrt{1+\frac{1}{4}{m}^2}$
$Ittouches\frac{x^2}{\frac{1}{4}}+\frac{y^2}{1}=1,e=\frac{\sqrt{3}}{2}$