Chord Length of Ellipse x²/4+y²/2=1 With Midpoint (1,½)

JEE Main 2025 — 23 January, Shift 2. Previous Year Question.

Problem

The length of the chord of the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{2}=1$, whose mid-point is $\left(1,\dfrac{1}{2}\right)$, is:

Key insight. Finding the two actual endpoints of the chord and subtracting them would mean solving for both intersection points explicitly. The standard “$T=S_1$” equation of a chord with a given midpoint skips that — it hands over the chord’s equation directly, and a touch of algebra with Vieta’s formulas gets the distance without ever solving for the endpoints individually.

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Approach

For a conic, the chord whose midpoint is $(x_1,y_1)$ has equation $T=S_1$, where $T$ is the tangent-style expression at $(x_1,y_1)$ and $S_1$ is the conic’s equation evaluated at that same point. Once this chord’s equation is known, substituting it into the ellipse’s equation gives a quadratic whose two roots are the $x$-coordinates of the chord’s endpoints — and the distance between the endpoints can be found from $(x_2-x_1)^2+(y_2-y_1)^2$ using only the sum and product of those roots, without solving the quadratic explicitly.

Solution

Step 1 — Write the chord’s equation using $T=S_1$

For the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{2}=1$ and midpoint $(x_1,y_1)=\left(1,\tfrac12\right)$:

$$T: \frac{x\,x_1}{4}+\frac{y\,y_1}{2} = S_1: \frac{x_1^2}{4}+\frac{y_1^2}{2}$$

Substituting $x_1=1,\ y_1=\tfrac12$:

$$\frac{x}{4}+\frac{y}{4} = \frac{1}{4}+\frac{1}{8}$$

Multiplying through by 4 and simplifying the right side to $\tfrac34$ (since $\tfrac14+\tfrac18=\tfrac38$, and multiplying by 4 gives $\tfrac32$ — carrying the multiplication through carefully):

$$x+y = \frac{3}{2}$$

Step 2 — Substitute into the ellipse equation

From the chord’s equation, $y = \dfrac{3}{2}-x$. Substitute into $\dfrac{x^2}{4}+\dfrac{y^2}{2}=1$:

$$\frac{x^2}{4}+\frac{\left(\frac32-x\right)^2}{2}=1$$

Expanding $\left(\tfrac32-x\right)^2 = \tfrac94 – 3x + x^2$:

$$\frac{x^2}{4}+\frac{9}{8}-\frac{3x}{2}+\frac{x^2}{2}=1$$

Combining the $x^2$ terms ($\tfrac14+\tfrac12=\tfrac34$) and moving everything to one side:

$$\frac{3x^2}{4}-\frac{3x}{2}+\frac{9}{8}-1=0 \implies \frac{3x^2}{4}-\frac{3x}{2}+\frac{1}{8}=0$$

Multiplying through by 8 to clear denominators:

$$6x^2-12x+1=0$$

Step 3 — Use Vieta’s formulas instead of solving for the roots

Let $x_1,x_2$ be the two roots of $6x^2-12x+1=0$. Then:

$$x_1+x_2 = \frac{12}{6}=2, \qquad x_1x_2=\frac{1}{6}$$

The quantity needed for the distance formula is $(x_2-x_1)^2$, which can be found without ever solving for the individual roots:

$$(x_2-x_1)^2 = (x_1+x_2)^2 – 4x_1x_2 = (2)^2 – 4\left(\frac16\right) = 4-\frac{2}{3}=\frac{10}{3}$$

Step 4 — Find $(y_2-y_1)^2$ from the chord’s equation

Since $y=\tfrac32-x$ along the chord, $y_2-y_1 = -(x_2-x_1)$, so:

$$(y_2-y_1)^2 = (x_2-x_1)^2 = \frac{10}{3}$$

Step 5 — Compute the chord length

$$\text{Length} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = \sqrt{\frac{10}{3}+\frac{10}{3}} = \sqrt{\frac{20}{3}}$$

Simplifying by rationalising:

$$\sqrt{\frac{20}{3}} = \frac{2\sqrt5}{\sqrt3} = \frac{2\sqrt5\cdot\sqrt3}{3} = \frac{2\sqrt{15}}{3}$$

Answer

Common mistakes

• Solving the quadratic $6x^2-12x+1=0$ directly for its roots. It works, but the numbers get messy fast; using $(x_2-x_1)^2=(x_1+x_2)^2-4x_1x_2$ from Vieta’s formulas avoids ever finding the individual roots.
• Forgetting that $y_2-y_1$ isn’t independent of $x_2-x_1$. Because the chord is a straight line $y=\tfrac32-x$, the two differences are related by a simple sign flip — recomputing $(y_2-y_1)^2$ from scratch as if it were unrelated wastes a step, though it gives the same answer if done correctly.

Practise next

• Find the length of the chord of the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ whose midpoint is $(1,1)$, using the same $T=S_1$ and Vieta approach.