JEE Main 2022 · 25 Jun · Shift 2 · Mathematics Q20

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

Answer and worked solution

Official answer: A

Intersection. $\frac{x^2}4+\frac{(x+1)^2}2=1$ gives $3x^2+4x-2=0$ : $x_1+x_2=-\frac43$ , $x_1x_2=-\frac23$ .

Chord. $(x_1-x_2)^2=\frac{16}9+\frac83=\frac{40}9$ , and on a line of slope 1, $PQ^2=2(x_1-x_2)^2=\frac{80}9$ .

Value. $r^2=\frac{20}9$ , so $(3r)^2=20$ .

(A)

Topics:Ellipse Distance and section formulae

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JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 20