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The line $\mathrm{y=x+1}$ 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 $\mathrm{\left (3r \right )^{2}}$ is equal to :
Option: 1
$20$

Option: 2
$12$

Option: 3
$11$

Option: 4
$8$

Answers (1)

best_answer

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

Since it lies o the point $\mathrm{\frac{x^{2}}{4}+\frac{y^{2}}{2}=1}$ we get

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

Here $\mathrm{\left(P_{1}-P_{2}\right)^{2}=\left(\frac{2 \sqrt{2}}{3}\right)^{2}-4(-2)=\frac{8}{9}+8=\frac{80}{9}}$

We know, $\mathrm{\left(P_{1}-P_{2}\right)^{2}=P Q^{2}}\\$

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

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

Hence the correct answer is option 4.

Posted by

Ritika Harsh

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