JEE Mathematics 2025 Question with Solution

Given: The ellipse is $\frac{x^2}{4} + \frac{y^2}{2} = 1$ and the midpoint of the chord is $\left(1, \frac{1}{2}\right)$.

Find: The length of the chord.

For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we have $a^2 = 4$ and $b^2 = 2$.

The chord of the ellipse having midpoint $\left(h,k\right)$ is given by

$$\frac{xh}{a^2} + \frac{yk}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2}.$$

Substituting $h = 1$ and $k = \frac{1}{2}$,

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

So,

$$2x + 2y = 3$$

or

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

Now solve this line with the ellipse. Put $y = \frac{3}{2} - x$ into

$$\frac{x^2}{4} + \frac{y^2}{2} = 1.$$

Then

$$\frac{x^2}{4} + \frac{\left(\frac{3}{2} - x\right)^2}{2} = 1.$$

This gives the two intersection points of the chord. The resulting distance between these two points simplifies to

$$\frac{2}{3}\sqrt{15}.$$

Therefore, the length of the chord is $\frac{2}{3}\sqrt{15}$, so the correct option is A.

Given: $\frac{x^2}{4} + \frac{y^2}{2} = 1$ and midpoint $\left(1, \frac{1}{2}\right)$.

Find: Length of the chord.

First identify

$$a = 2, \qquad b = \sqrt{2}.$$

For an ellipse, the chord with midpoint $\left(h,k\right)$ is obtained by the midpoint form. Using $h = 1$ and $k = \frac{1}{2}$,

$$\frac{xh}{a^2} + \frac{yk}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2}$$

becomes

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

Hence the chord is

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

Now intersect this line with the ellipse. From

$$y = \frac{3}{2} - x,$$

substitute into the ellipse:

$$\frac{x^2}{4} + \frac{\left(\frac{3}{2} - x\right)^2}{2} = 1.$$

Expand:

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

So,

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

Multiply by $8$:

$$2x^2 + 4x^2 - 12x + 9 = 8.$$

Thus,

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

The roots are

$$x = \frac{12 \pm \sqrt{144 - 24}}{12} = \frac{12 \pm 2\sqrt{30}}{12} = 1 \pm \frac{\sqrt{30}}{6}.$$

Corresponding values of $y$ are

$$y = \frac{3}{2} - x = \frac{1}{2} \mp \frac{\sqrt{30}}{6}.$$

So the endpoints are

$$\left(1 + \frac{\sqrt{30}}{6}, \frac{1}{2} - \frac{\sqrt{30}}{6}\right)$$

and

$$\left(1 - \frac{\sqrt{30}}{6}, \frac{1}{2} + \frac{\sqrt{30}}{6}\right).$$

Now use the distance formula:

$$L = \sqrt{\left(\frac{\sqrt{30}}{3}\right)^2 + \left(-\frac{\sqrt{30}}{3}\right)^2}$$ $$= \sqrt{\frac{30}{9} + \frac{30}{9}}$$ $$= \sqrt{\frac{60}{9}} = \sqrt{\frac{20}{3}} = \frac{2\sqrt{15}}{3}.$$

Therefore, the length of the chord is $\frac{2}{3}\sqrt{15}$.