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  • The lie meets the ellipse <img alt="\mathrm{\frac{x^{2}}{4}+\frac{y^{2}}{2}=1}" sr

The lie \mathrm{y=x+1} meets the ellipse \mathrm{\frac{x^{2}}{4}+\frac{y^{2}}{2}=1} at two points P and Q. If \mathrm{r} is the radius of the circle with PQ as diameter then \mathrm{\left (3r^{2} \right )} is equal to :

Option: 1

20


Option: 2

12


Option: 3

11


Option: 4

8


Answers (1)

best_answer

Let P be the point  \mathrm{\left ( -1+\frac{P}{\sqrt{2}},\frac{P}{\sqrt{2}} \right )}

 Since it lies on the ellipse  \mathrm{\frac{x^{2}}{4}+\frac{y^{2}}{2}=1} , we get

\mathrm{\frac{\left(-1+\frac{p}{\sqrt{2}}\right)^{2}}{4}+\frac{\left(\frac{p}{\sqrt{2}}\right)^{2}}{2}=1 \Rightarrow \frac{3 p^{2}}{8}-\frac{p}{2 \sqrt{2}}-\frac{3}{4}=0}\\

\mathrm{Here \left(P_{1}-P_{2}\right)^{2}=\left(\frac{2 \sqrt{2}}{3}\right)^{2}-4(-2)=\frac{8}{9}+8=\frac{80}{9}} \mathrm{we \: know, \left(P_{1}-P_{2}\right)^{2}=P Q^{2}}\\

\mathrm{\Rightarrow P Q^{2}=\frac{80}{9}}\\

\mathrm{Now \: r=\frac{P Q}{2} \Rightarrow 9r^{2}=\frac{9(P Q)^{2}}{4}=\frac{9}{4} \times \frac{80}{9}=20}

 Hence the correct answer is option 1

Posted by

manish painkra

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