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  • The line intersects the ellipse <img alt="\mathrm{\frac{x^2}{4}+y^2=1}" src="https://entra

The line \mathrm{x=2 y}intersects the ellipse \mathrm{\frac{x^2}{4}+y^2=1} at the point P and Q. The equation of the circle with PQ as diameter is

Option: 1

\mathrm{x^2+y^2=\frac{1}{2}}


Option: 2

\mathrm{x^2+y^2=1}


Option: 3

\mathrm{x^2+y^2=2}


Option: 4

\mathrm{x^2+y^2=\frac{5}{2}}


Answers (1)

best_answer

\mathrm{x=2 y}                              ....(i)

\mathrm{\text { and } \frac{x^2}{4}+y^2=1}           ....(ii)

On solving, \mathrm{2 y^2=1, y= \pm \frac{1}{\sqrt{2}} \Rightarrow x= \pm \sqrt{2}}

\mathrm{ \therefore P\left(\sqrt{2}, \frac{1}{\sqrt{2}}\right) \text { and } Q\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right) \text { (say) } }

\mathrm{\therefore \quad} Circle with PQ as diameter is

\mathrm{ \begin{aligned} & (x-\sqrt{2})(x+\sqrt{2})+\left(y-\frac{1}{\sqrt{2}}\right)\left(y+\frac{1}{\sqrt{2}}\right)=0 \\ & \Rightarrow x^2+y^2=\frac{5}{2} \end{aligned} }

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vishal kumar

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