The line y=x+1 meets the ellipse frac{x^{2}}{ | vedclass

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Question

Ellipse $x^{2}+2 y^{2}=4$

Line $y=x+1$

Point of intersection

$x^{2}+2(x+1)^{2}=4$

$3 x^{2}+4 x-2=0$

$\left|x_{1}-x_{2}\right|=\frac{\sq

Vedclass · Accepted Answer

$20$

Vedclass · Answer

Ellipse $x^{2}+2 y^{2}=4$

Line $y=x+1$

Point of intersection

$x^{2}+2(x+1)^{2}=4$

$3 x^{2}+4 x-2=0$

$\left|x_{1}-x_{2}ight|=\frac{\sqrt{40}}{3}$

$AB =2 r =\left| x _{1}- x _{2}ight| \sqrt{1+ m ^{2}}$

$m$ is slope of given line

$AB =\frac{\sqrt{40}}{3} \sqrt{1+1}$

$2 r =\frac{\sqrt{80}}{3} \Rightarrow r =\frac{\sqrt{80}}{6}$

$(3 r )^{2}=\left(3 	imes \frac{\sqrt{80}}{6}ight)^{2}=\frac{80}{4}=20$

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

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