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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}


Answers (1)

best_answer

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}.}

Posted by

Shailly goel

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