JEE Main 2025 — 28 January, Shift 2. Previous Year Question.
Problem
If the midpoint of a chord of the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ is $\left(\sqrt{2}, \dfrac{4}{3}\right)$, and the length of the chord is $\dfrac{2\sqrt{\alpha}}{3}$, then $\alpha$ is:
Key insight. The equation of a chord with a given midpoint has a ready-made shortcut — the “$T = S_1$” formula — which skips straight past finding the chord’s slope from scratch. Once you have the chord’s equation, the two endpoints fall out of solving it simultaneously with the ellipse.
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Approach
Use the $T = S_1$ formula to write down the chord’s equation directly from its midpoint, without needing calculus or a slope calculation. Then solve that line simultaneously with the ellipse to find the two actual endpoints, and finish with the distance formula.
Solution
Step 1 — Write the chord’s equation using $T = S_1$
For the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ with chord midpoint $(x_1, y_1) = \left(\sqrt{2}, \dfrac{4}{3}\right)$, the chord’s equation is:
$$\frac{xx_1}{9} + \frac{yy_1}{4} = \frac{x_1^2}{9} + \frac{y_1^2}{4}$$
Substituting $x_1 = \sqrt{2}$, $y_1 = \dfrac{4}{3}$:
$$\frac{\sqrt{2}\,x}{9} + \frac{y}{3} = \frac{2}{9} + \frac{4}{9} = \frac{6}{9} = \frac{2}{3}$$
Multiplying through by 9:
$$\sqrt{2}\,x + 3y = 6$$
Step 2 — Solve the chord and the ellipse together
From the chord’s equation, $y = \dfrac{6 – \sqrt{2}x}{3}$. Substituting into $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ and multiplying through by 36:
$$4x^2 + (6 – \sqrt{2}x)^2 = 36$$
$$4x^2 + 36 – 12\sqrt{2}x + 2x^2 = 36$$
$$6x^2 – 12\sqrt{2}x = 0 \implies 6x(x – 2\sqrt{2}) = 0$$
So $x = 0$ or $x = 2\sqrt{2}$.
Step 3 — Find the two endpoints
- At $x = 0$: $y = \dfrac{6-0}{3} = 2$, giving the point $(0, 2)$.
- At $x = 2\sqrt{2}$: $y = \dfrac{6 – \sqrt{2}\cdot 2\sqrt{2}}{3} = \dfrac{6-4}{3} = \dfrac{2}{3}$, giving the point $\left(2\sqrt{2}, \dfrac{2}{3}\right)$.
Step 4 — Compute the chord length
$$\text{length} = \sqrt{(2\sqrt{2} – 0)^2 + \left(\frac{2}{3} – 2\right)^2} = \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{72+16}{9}} = \sqrt{\frac{88}{9}} = \frac{2\sqrt{22}}{3}$$
Step 5 — Match against the given form
Comparing $\dfrac{2\sqrt{22}}{3}$ with $\dfrac{2\sqrt{\alpha}}{3}$:
$$\alpha = 22$$
Answer
$$\alpha = 22$$
Common mistakes
- Trying to find the chord’s slope by differentiating implicitly instead of using $T = S_1$. It works, but the midpoint-chord formula is built exactly for this situation and avoids an extra layer of algebra.
- Arithmetic slips squaring $(6-\sqrt{2}x)$. The cross term $-12\sqrt{2}x$ is easy to drop; keeping the $\sqrt{2}$ explicit through each line avoids losing it.
Practise next
- Find the length of the chord of $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ whose midpoint is $(1, 1)$, using the same $T=S_1$ approach.
Show answer
$\dfrac{\sqrt{40103}}{30}\approx6.68$. $T=S_1$ gives $\dfrac{x}{16}+\dfrac{y}{9}=\dfrac{1}{16}+\dfrac{1}{9}=\dfrac{25}{144}$, that is $9x+16y=25$.
Substituting into $9x^2+16y^2=144$ leaves $400y^2-800y-671=0$, whose roots sum to $2$ — confirming the midpoint — and multiply to $-\tfrac{671}{400}$, so $(y_1-y_2)^2=\dfrac{1071}{100}$.
From the chord, $x_1-x_2=-\tfrac{16}{9}(y_1-y_2)$, so the squared length is $\left(\tfrac{256}{81}+1\right)\tfrac{1071}{100}=\dfrac{40103}{900}$. This one does not land on a tidy surd — the method is what carries over.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.