Let x=9 be a directrix of an ellipse E, whose centre is at the origin and eccentricity is 1/3. 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:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q12 – JEE-MAIN 2026

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Home Maths Conic SectionsQuestion #121

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 AB be a chord passing through P. Then the locus of the mid point of AB is:

Clue 1 of 3: Ellipse Properties

Recall the standard equation of an ellipse and the relationships between its parameters (center, directrix, eccentricity, foci).

🧠 Full Solution Path

Step 1: Determine Ellipse Parameters and Focus✦ Active

Given the directrix $x = 9$ , center at origin, and eccentricity $e = \frac{1}{3}$ . For an ellipse with center at origin, the directrix is $x = \frac{a}{e}$ . Thus, $\frac{a}{e} = 9 \Rightarrow a = 9 e = 9 \times \frac{1}{3} = 3$ . The relation between $a, b, e$ is $b^{2} = a^{2} (1 - e^{2})$ . Substituting the values, $b^{2} = 3^{2} (1 - {(\frac{1}{3})}^{2}) = 9 (1 - \frac{1}{9}) = 9 (\frac{8}{9}) = 8$ . The equation of the ellipse is $\frac{x^{2}}{9} + \frac{y^{2}}{8} = 1$ . The foci are at $(\pm a e, 0)$ . So, $a e = 3 \times \frac{1}{3} = 1$ . Since $P (α, 0)$ is a focus and $α > 0$ , we have $P (1, 0)$ .

Step 2: Apply Chord Midpoint Formula○ Expand

Let the midpoint of the chord $A B$ be $M (h, k)$ . The equation of the chord of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with midpoint $(h, k)$ is given by $T = S_{1}$ , which is:

\frac{x h}{a^{2}} + \frac{y k}{b^{2}} = \frac{h^{2}}{a^{2}} + \frac{k^{2}}{b^{2}}

Since the chord passes through the focus $P (1, 0)$ , we substitute $(x, y) = (1, 0)$ into the chord equation:

\frac{(1) h}{a^{2}} + \frac{(0) k}{b^{2}} = \frac{h^{2}}{a^{2}} + \frac{k^{2}}{b^{2}}

This simplifies to:

\frac{h}{a^{2}} = \frac{h^{2}}{a^{2}} + \frac{k^{2}}{b^{2}}

💡 Teacher's Secret Hint

Remember the $T = S_{1}$ formula for the chord with a given midpoint. It's a common shortcut for conic sections.

Step 3: Substitute Parameters and Find Locus○ Expand

Substitute $a^{2} = 9$ and $b^{2} = 8$ into the equation from Step 2:

\frac{h}{9} = \frac{h^{2}}{9} + \frac{k^{2}}{8}

To find the locus, replace $(h, k)$ with $(x, y)$ :

\frac{x}{9} = \frac{x^{2}}{9} + \frac{y^{2}}{8}

Multiply the entire equation by $72$ (the LCM of $9$ and $8$ ) to clear the denominators:

8 x = 8 x^{2} + 9 y^{2}

Rearrange the terms to match the given options:

9 y^{2} = 8 x - 8 x^{2}

Factor out $8 x$ from the right side:

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

This matches option 1.

💡 Teacher's Secret Hint

Always simplify the equation to its most standard form and compare it carefully with the given options.

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Maths Question 12 – JEE-MAIN 2026

Clue 1 of 3: Ellipse Properties

🧠 Full Solution Path