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10-2. Parabola, Ellipse, Hyperbola
hard
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
A
$20$
B
$12$
C
$11$
D
$8$
(JEE MAIN-2022)
Solution
Ellipse $x^{2}+2 y^{2}=4$
Line $y=x+1$
Point of intersection
$x^{2}+2(x+1)^{2}=4$
$3 x^{2}+4 x-2=0$
$\left|x_{1}-x_{2}\right|=\frac{\sqrt{40}}{3}$
$AB =2 r =\left| x _{1}- x _{2}\right| \sqrt{1+ m ^{2}}$
$m$ is slope of given line
$AB =\frac{\sqrt{40}}{3} \sqrt{1+1}$
$2 r =\frac{\sqrt{80}}{3} \Rightarrow r =\frac{\sqrt{80}}{6}$
$(3 r )^{2}=\left(3 \times \frac{\sqrt{80}}{6}\right)^{2}=\frac{80}{4}=20$
Standard 11
Mathematics
