JEE Main 2022 · 26 Jun · Shift 2 · Mathematics Q12

MathematicsSingle correct+4 / -1Question 12 of 90

The locus of the mid point of the line segment joining the point $(4,3)$ and the points on the ellipse $x^2+2y^2=4$ is an ellipse with eccentricity :

\dfrac{\sqrt3}2

\dfrac1{2\sqrt2}

\dfrac1{\sqrt2}

\dfrac12

Answer and worked solution

Official answer: C

Midpoint $(h,k)$ . The ellipse point is $(2h-4,2k-3)$ : $(2h-4)^2+2(2k-3)^2=4$ , i.e. $(h-2)^2+2\left(k-\frac32\right)^2=1$ .

Shape. $a^2=1$ , $b^2=\frac12$ : $e=\sqrt{1-\frac12}=\dfrac1{\sqrt2}$ .

(C)

Topics:Ellipse Locus

JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 12