JEE Mathematics 2025 Question with Solution

Given: The ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$. The midpoint of the chord is $\left( \sqrt{2}, \frac{4}{3} \right)$ and the chord length is $\frac{2\sqrt{\alpha}}{3}$.

Find: The value of $\alpha$.

For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the chord whose midpoint is $\left(x_1,y_1\right)$ has equation

$$\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}$$

Here, $a^2 = 9,\; b^2 = 4,\; x_1 = \sqrt{2},\; y_1 = \frac{4}{3}$. So

$$\frac{x\sqrt{2}}{9} + \frac{y\left(\frac{4}{3}\right)}{4} = \frac{\left(\sqrt{2}\right)^2}{9} + \frac{\left(\frac{4}{3}\right)^2}{4}$$ $$\frac{\sqrt{2}x}{9} + \frac{y}{3} = \frac{2}{9} + \frac{4}{9} = \frac{2}{3}$$

Hence, the chord is

$$\sqrt{2}x + 3y = 6$$

Now put $y = \frac{6 - \sqrt{2}x}{3}$ in the ellipse:

$$\frac{x^2}{9} + \frac{1}{4}\left(\frac{6 - \sqrt{2}x}{3}\right)^2 = 1$$ $$\frac{x^2}{9} + \frac{1}{36}(36 - 12\sqrt{2}x + 2x^2) = 1$$

Multiplying by $36$,

$$4x^2 + 36 - 12\sqrt{2}x + 2x^2 = 36$$ $$6x^2 - 12\sqrt{2}x = 0$$ $$6x(x - 2\sqrt{2}) = 0$$

So, $x = 0$ or $x = 2\sqrt{2}$.

Corresponding values of $y$ are:

• for $x = 0$, $y = 2$
• for $x = 2\sqrt{2}$, $y = \frac{2}{3}$

Thus the endpoints are $A(0,2)$ and $B\left(2\sqrt{2}, \frac{2}{3}\right)$. Using the distance formula,

$$AB = \sqrt{(2\sqrt{2} - 0)^2 + \left(\frac{2}{3} - 2\right)^2}$$ $$AB = \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{88}{9}} = \frac{2\sqrt{22}}{3}$$

Given that the length is $\frac{2\sqrt{\alpha}}{3}$, we get $\alpha = 22$.

Therefore, the correct option is B.

Given: Midpoint of the chord is $\left( \sqrt{2}, \frac{4}{3} \right)$ in the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

Find: The value of $\alpha$ when chord length is $\frac{2\sqrt{\alpha}}{3}$.

Write the ellipse in standard form with $a = 3$ and $b = 2$. Using the midpoint form of a chord of an ellipse:

$$\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}$$

Substitute $\left(x_1,y_1\right) = \left(\sqrt{2}, \frac{4}{3}\right)$:

$$\frac{x\sqrt{2}}{9} + \frac{y\left(\frac{4}{3}\right)}{4} = \frac{2}{9} + \frac{16}{36}$$ $$\frac{\sqrt{2}x}{9} + \frac{y}{3} = \frac{2}{9} + \frac{4}{9} = \frac{2}{3}$$ $$\sqrt{2}x + 3y = 6$$

Substitute $y = \frac{6 - \sqrt{2}x}{3}$ into the ellipse equation:

$$\frac{x^2}{9} + \frac{1}{4}\left(\frac{6 - \sqrt{2}x}{3}\right)^2 = 1$$ $$\frac{x^2}{9} + \frac{1}{36}(6 - \sqrt{2}x)^2 = 1$$ $$\frac{x^2}{9} + \frac{1}{36}(36 - 12\sqrt{2}x + 2x^2) = 1$$ $$4x^2 + 36 - 12\sqrt{2}x + 2x^2 = 36$$ $$6x^2 - 12\sqrt{2}x = 0$$ $$6x(x - 2\sqrt{2}) = 0$$

Hence the points of intersection are obtained from $x = 0$ and $x = 2\sqrt{2}$. Then

$$x = 0 \Rightarrow y = 2$$ $$x = 2\sqrt{2} \Rightarrow y = \frac{2}{3}$$

So endpoints are $A(0,2)$ and $B\left(2\sqrt{2}, \frac{2}{3}\right)$. Now calculate the length:

$$AB = \sqrt{(2\sqrt{2})^2 + \left(-\frac{4}{3}\right)^2}$$ $$AB = \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{88}{9}} = \frac{2\sqrt{22}}{3}$$

Comparing with $\frac{2\sqrt{\alpha}}{3}$ gives $\alpha = 22$.

Therefore, the answer is $22$ and the correct option is B.