Ellipse – Chord with given Mid-point – JEE Main 2025 Shift 2

Solution:

Let the equation of the chord with mid-point $(x_1, y_1)$ be $T = S_1$.
Given point: $(x_1, y_1) = (1, \frac{1}{2})$.
Equation of Ellipse: $\frac{x^2}{4} + \frac{y^2}{2} – 1 = 0$.

Applying $T = S_1$:
$$\frac{x(1)}{4} + \frac{y(1/2)}{2} – 1 = \frac{(1)^2}{4} + \frac{(1/2)^2}{2} – 1$$
$$\frac{x}{4} + \frac{y}{4} = \frac{1}{4} + \frac{1}{8}$$
$$\frac{x+y}{4} = \frac{3}{8}$$
$$x + y = \frac{3}{2} \Rightarrow y = \frac{3}{2} – x$$

To find the length of the chord, we find the points of intersection with the ellipse.
Substitute $y = \frac{3}{2} – x$ into the ellipse equation $x^2 + 2y^2 = 4$:
$$x^2 + 2\left(\frac{3}{2} – x\right)^2 = 4$$
$$x^2 + 2\left(\frac{9}{4} – 3x + x^2\right) = 4$$
$$x^2 + \frac{9}{2} – 6x + 2x^2 = 4$$
$$3x^2 – 6x + \frac{1}{2} = 0$$
$$6x^2 – 12x + 1 = 0$$

Let roots be $x_A$ and $x_B$.
Sum of roots ($x_A + x_B$) = $\frac{12}{6} = 2$.
Product of roots ($x_A x_B$) = $\frac{1}{6}$.
Difference of roots $|x_A – x_B| = \sqrt{(x_A+x_B)^2 – 4x_Ax_B}$:
$$|x_A – x_B| = \sqrt{(2)^2 – 4(\frac{1}{6})} = \sqrt{4 – \frac{2}{3}} = \sqrt{\frac{10}{3}}$$

The length of the chord $L$ is given by distance formula. Since the line has slope $m = -1$:
$$L = \sqrt{(x_A – x_B)^2 + (y_A – y_B)^2}$$
$$L = |x_A – x_B| \sqrt{1 + m^2}$$
$$L = \sqrt{\frac{10}{3}} \cdot \sqrt{1 + (-1)^2}$$
$$L = \sqrt{\frac{10}{3}} \cdot \sqrt{2} = \sqrt{\frac{20}{3}}$$
$$L = 2\sqrt{\frac{5}{3}} = \frac{2\sqrt{15}}{3}$$

Ans. (1)