JEE Main 2026 Apr 6 Shift 2, Mathematics Q12: Ellipse: Chord, Tangent and Normal

JEE Main2026Apr 6, Shift 2Mathematics

Let $x = 9$ be a directrix of an ellipse $E$ , whose centre is at the origin and eccentricity is $\frac{1}{3}$ . Let $P (α, 0)$ , $α > 0$ , be a focus of $E$ and $A B$ be a chord passing through $P$ . Then the locus of the mid point of $A B$ is:

A
$9 y^{2} = 8 x (1 - x)$
B
$3 y^{2} = 4 x (1 - x)$
C
$9 y^{2} = 8 x (x - 1)$
D
$3 y^{2} = 4 x (x - 1)$

Solution

From $\frac{a}{e} = 9$ with $e = \frac{1}{3}$ : $a = 3$ , and $b^{2} = a^{2} (1 - e^{2}) = 9 \cdot \frac{8}{9} = 8$ . So $E : \frac{x ^{2}}{9} + \frac{y ^{2}}{8} = 1$ .

Focus $α = a e = 1$ , so $P = (1, 0)$ .

For a chord with midpoint $(h, k)$ , the chord equation is $T = S_{1}$ :

$\frac{h x}{9} + \frac{k y}{8} = \frac{h ^{2}}{9} + \frac{k ^{2}}{8}$ .

Substitute $(1, 0)$ (since the chord passes through $P$ ):

$\frac{h}{9} = \frac{h ^{2}}{9} + \frac{k ^{2}}{8}$ $\Rightarrow$ $\frac{k ^{2}}{8} = \frac{h - h ^{2}}{9}$ $\Rightarrow$ $9 k^{2} = 8 h (1 - h)$ .

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

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q12: Ellipse: Chord, Tangent and Normal

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