Mathematics · Ellipse

JEE Main 2025 — 23 January, Evening Shift — Question 15

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

Answer: A

\text{Given\; ellipse:\; } \frac{x^2}{4} + \frac{y^2}{2} = 1

\text{Midpoint\; of\; chord:\; } (x_1, y_1) = (1, \tfrac{1}{2})

\text{Equation\; of\; chord\; with\; midpoint\; form:\; } \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}

\Rightarrow \frac{x(1)}{4} + \frac{y(\frac{1}{2})}{2} = \frac{1^2}{4} + \frac{(\frac{1}{2})^2}{2}

\Rightarrow \frac{x}{4} + \frac{y}{4} = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}

\Rightarrow 2x + 2y = 3 \quad \text{or} \quad y = \frac{3 - 2x}{2}

\text{Substitute\; in\; ellipse:\; } \frac{x^2}{4} + \frac{1}{2}\left( \frac{3 - 2x}{2} \right)^2 = 1

\Rightarrow \frac{x^2}{4} + \frac{9 - 12x + 4x^2}{8} = 1

\Rightarrow 2x^2 + 9 - 12x + 4x^2 = 8

\Rightarrow 6x^2 - 12x + 1 = 0

x = \frac{12 \pm \sqrt{(-12)^2 - 4(6)(1)}}{12}

x = \frac{12 \pm \sqrt{144 - 24}}{12} = \frac{12 \pm \sqrt{120}}{12} = \frac{12 \pm 2\sqrt{30}}{12}

\Rightarrow x = 1 \pm \frac{\sqrt{30}}{6}

y = \frac{3 - 2x}{2} = \frac{3 - 2(1 \pm \frac{\sqrt{30}}{6})}{2} = \frac{1 \mp \frac{\sqrt{30}}{3}}{2} = \frac{1}{2} \mp \frac{\sqrt{30}}{6}

\text{Endpoints:\; } A\left(1 + \frac{\sqrt{30}}{6}, \frac{1}{2} - \frac{\sqrt{30}}{6}\right), \quad B\left(1 - \frac{\sqrt{30}}{6}, \frac{1}{2} + \frac{\sqrt{30}}{6}\right)

AB = \sqrt{\left(2 \times \frac{\sqrt{30}}{6}\right)^2 + \left(2 \times \frac{\sqrt{30}}{6}\right)^2}

= \sqrt{2 \left(\frac{2\sqrt{30}}{6}\right)^2} = \sqrt{2 \cdot \frac{4 \times 30}{36}} = \sqrt{\frac{240}{36}} = \sqrt{\frac{20}{3}} = \frac{2\sqrt{15}}{3}

\boxed{\text{Length\; of\; chord\; } = \frac{2\sqrt{15}}{3}}

Answer key and solution verified before publishing.

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