JEE Main 2026 · 6 Apr · Shift 2 · Mathematics Q12

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MathematicsSingle correct+4 / -1Question 12 of 75

Let $x=9$ be a directrix of an ellipse E, whose centre is at the origin and eccentricity is $\dfrac13$ . Let $P(\alpha,0)$ , $\alpha>0$ , be a focus of E and AB be a chord passing through P. Then the locus of the mid point of AB is :

9y^2=8x(1-x)

3y^2=4x(1-x)

9y^2=8x(x-1)

3y^2=4x(x-1)

Answer and worked solution

Official answer: A

Ellipse. $\dfrac ae=9$ and $e=\tfrac13$ give $a=3$ , focus $(1,0)$ , $b^2=8$ .

Midpoint (h, k) of a chord. The chord is $\dfrac{hx}9+\dfrac{ky}8=\dfrac{h^2}9+\dfrac{k^2}8$ . Through $(1,0)$ :
$\frac h9=\frac{h^2}9+\frac{k^2}8\ \Rightarrow\ 9k^2=8h(1-h).$

Locus: $9y^2=8x(1-x)$ .

(A)

Topics:Ellipse Locus

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JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 12