αx+βy=109 Is Chord of x²/9+y²/4=1 With Midpoint (5/2,1/2), Find α+β

JEE Main 2025 — 29 January, Evening Shift. Previous Year Question.

Problem

If $\alpha x + \beta y = 109$ is the equation of the chord of the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ whose midpoint is $\left(\dfrac{5}{2},\dfrac{1}{2}\right)$, then $\alpha + \beta$ is equal to:

Key insight. There’s a direct formula for the chord of a conic with a given midpoint — it’s the “$T=S_1$” equation, obtained by replacing $x^2 \to xx_1$, $y^2\to yy_1$ in the conic’s equation. Once that equation is written out, matching its coefficients against the given $\alpha x+\beta y=109$ hands over $\alpha$ and $\beta$ directly, with no need to find the chord’s endpoints.

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Approach

Write the $T=S_1$ equation for the ellipse using the given midpoint $(x_1,y_1)=\left(\frac{5}{2},\frac{1}{2}\right)$, simplify it into the standard linear form $Ax+By=C$, then scale the whole equation so its constant term matches $109$ — at that point, the coefficients of $x$ and $y$ are $\alpha$ and $\beta$.

Solution

Step 1 — Write the T = S₁ equation

For the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$, the chord with midpoint $(x_1,y_1)$ has equation:

$$\frac{xx_1}{9}+\frac{yy_1}{4} = \frac{x_1^2}{9}+\frac{y_1^2}{4}$$

Step 2 — Substitute the midpoint

With $x_1=\dfrac{5}{2}$, $y_1=\dfrac{1}{2}$:

$$\frac{x\cdot\frac{5}{2}}{9}+\frac{y\cdot\frac{1}{2}}{4} = \frac{\left(\frac{5}{2}\right)^2}{9}+\frac{\left(\frac{1}{2}\right)^2}{4}$$

$$\frac{5x}{18}+\frac{y}{8} = \frac{25}{36}+\frac{1}{16}$$

Step 3 — Simplify the right-hand side

Using a common denominator of $144$:

$$\frac{25}{36} = \frac{100}{144}, \qquad \frac{1}{16} = \frac{9}{144}$$

$$\frac{25}{36}+\frac{1}{16} = \frac{109}{144}$$

Step 4 — Simplify the left-hand side to a matching form

Using a common denominator of $72$ on the left:

$$\frac{5x}{18}+\frac{y}{8} = \frac{20x+9y}{72}$$

So the equation is:

$$\frac{20x+9y}{72} = \frac{109}{144}$$

Step 5 — Scale so the constant becomes 109

Multiplying both sides by $144$:

$$2(20x+9y) = 109 \implies 40x+18y = 109$$

Step 6 — Match coefficients

Comparing with $\alpha x+\beta y=109$:

$$\alpha = 40, \qquad \beta = 18 \implies \alpha+\beta = 58$$

Answer

Common mistakes

• Trying to find the actual endpoints of the chord first. This works but requires solving the ellipse and a line simultaneously — a much longer route than the direct $T=S_1$ formula, which needs only the midpoint.
• Forgetting to scale the equation so the constant term is exactly $109$ before reading off $\alpha$ and $\beta$. Reading coefficients off an unscaled equation like $\dfrac{20x+9y}{72}=\dfrac{109}{144}$ directly would give the wrong values.

Practise next

• Find $\alpha+\beta$ if $\alpha x+\beta y = 50$ is the chord of $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ with midpoint $(2,1)$, using the same $T=S_1$ method.