JEE Main 2025. Previous Year Question.
Problem
The equation of the chord of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$, whose midpoint is $(3,1)$, is:
Key insight. The chord of any conic with a given midpoint is found directly using the “$T=S_1$” formula — replace $x^2\to xx_1$ and $y^2\to yy_1$ in the conic’s equation, using the midpoint’s coordinates for $x_1,y_1$. No need to find the chord’s actual endpoints.
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Approach
Write the $T=S_1$ equation for the ellipse using the given midpoint $(x_1,y_1)=(3,1)$, then simplify by clearing denominators to get the chord’s equation in standard linear form.
Solution
Step 1 — Write the T = S₁ equation
For the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$, the chord with midpoint $(x_1,y_1)$ has equation:
$$\frac{xx_1}{25}+\frac{yy_1}{16} = \frac{x_1^2}{25}+\frac{y_1^2}{16}$$
Step 2 — Substitute the midpoint (3,1)
$$\frac{3x}{25}+\frac{y}{16} = \frac{9}{25}+\frac{1}{16}$$
Step 3 — Clear denominators using the LCM of 25 and 16
The LCM of $25$ and $16$ is $400$. Multiplying both sides by $400$:
$$400\left(\frac{3x}{25}\right)+400\left(\frac{y}{16}\right) = 400\left(\frac{9}{25}\right)+400\left(\frac{1}{16}\right)$$
$$16(3x)+25(y) = 16(9)+25(1)$$
Step 4 — Simplify
$$48x+25y = 144+25 = 169$$
Answer
$$48x+25y=169$$
Common mistakes
- Forgetting to multiply by the correct LCM. Using an incorrect common denominator (or forgetting to convert $\frac{3x}{25}$ and $\frac{y}{16}$ to whole-number coefficients consistently) leads to a chord equation that doesn’t match any of the given options.
- Confusing this with the tangent equation. The $T=S_1$ chord-of-midpoint formula looks identical in structure to the tangent equation $T=0$, but the right-hand side ($S_1$, not $0$) is what makes it a chord through an interior point rather than a tangent line.
Practise next
- Find the equation of the chord of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ whose midpoint is $(2,1)$, using the same $T=S_1$ method.
Show answer
$9x+8y=26$. The chord with a given midpoint satisfies $T=S_1$, where both sides are built from the midpoint $(2,1)$.
$$\frac{2x}{16}+\frac{y}{9}=\frac{4}{16}+\frac{1}{9}\quad\Longrightarrow\quad\frac{x}{8}+\frac{y}{9}=\frac{13}{36}.$$
Multiplying through by $72$ gives $9x+8y=26$, and substituting the midpoint confirms it: $18+8=26$. No intersection points ever have to be found.
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