JEE Main 2026 — Mathematics Coordinate Geometry

Given the directrix of the ellipse is $x = \dfrac{a}{e} = 9$x=ea​=9 and eccentricity $e = \dfrac{1}{3}$e=31​.

$$\dfrac{a}{1/3} = 9 \Rightarrow a = 3$$1/3a​=9⇒a=3

The value of $b^2$b2 is given by $b^2 = a^2(1 - e^2) = 9\left(1 - \dfrac{1}{9}\right) = 8$b2=a2(1−e2)=9(1−91​)=8.

The equation of the ellipse is $\dfrac{x^2}{9} + \dfrac{y^2}{8} = 1$9x2​+8y2​=1.

The focus $P(\alpha, 0)$P(α,0) for $\alpha > 0$α>0 is at $(ae, 0) = \left(3 \times \dfrac{1}{3}, 0\right) = (1, 0)$(ae,0)=(3×31​,0)=(1,0).

Let the midpoint of the chord $AB$AB be $(h, k)$(h,k). The equation of the chord in terms of its midpoint is given by $T = S_1$T=S1​ :

$$\dfrac{hx}{9} + \dfrac{ky}{8} = \dfrac{h^2}{9} + \dfrac{k^2}{8}$$9hx​+8ky​=9h2​+8k2​

Since the chord passes through the focus $P(1, 0)$P(1,0), substituting $x = 1$x=1 and $y = 0$y=0 gives:

$$\dfrac{h}{9} = \dfrac{h^2}{9} + \dfrac{k^2}{8}$$9h​=9h2​+8k2​

$$\dfrac{k^2}{8} = \dfrac{h}{9} - \dfrac{h^2}{9}$$8k2​=9h​−9h2​

$$9k^2 = 8h(1 - h)$$9k2=8h(1−h)

Replacing $(h, k)$(h,k) with $(x, y)$(x,y), the locus of the midpoint is:

$$9y^2 = 8x(1 - x)$$9y2=8x(1−x)

Answer: $9y^2 = 8x(1-x)$9y2=8x(1−x)