Coordinate Geometry – Ellipse – JEE Main 28 Jan 2025 Shift 2

Solution:

Let the midpoint of the chord be $M(\sqrt{2}, \frac{4}{3})$.
The equation of the chord bisected at a given point is $T = S_1$.

Equation of Ellipse $S \equiv \frac{x^2}{9} + \frac{y^2}{4} – 1 = 0$.

Using $T = S_1$:
$$\frac{x(\sqrt{2})}{9} + \frac{y(4/3)}{4} – 1 = \frac{(\sqrt{2})^2}{9} + \frac{(4/3)^2}{4} – 1$$

$$\frac{\sqrt{2}x}{9} + \frac{y}{3} = \frac{2}{9} + \frac{16/9}{4}$$
$$\frac{\sqrt{2}x}{9} + \frac{3y}{9} = \frac{2}{9} + \frac{4}{9}$$
$$\frac{\sqrt{2}x + 3y}{9} = \frac{6}{9}$$
$$\sqrt{2}x + 3y = 6 \Rightarrow y = \frac{6-\sqrt{2}x}{3}$$

To find the points of intersection, substitute $y$ into the ellipse equation:
$$4x^2 + 9y^2 = 36$$
$$4x^2 + 9\left(\frac{6-\sqrt{2}x}{3}\right)^2 = 36$$
$$4x^2 + (6-\sqrt{2}x)^2 = 36$$
$$4x^2 + (36 + 2x^2 – 12\sqrt{2}x) = 36$$
$$6x^2 – 12\sqrt{2}x = 0$$
$$6x(x – 2\sqrt{2}) = 0$$

So, $x_1 = 0$ and $x_2 = 2\sqrt{2}$.

If $x_1 = 0$, $y_1 = \frac{6-0}{3} = 2$. Point $A(0, 2)$.

If $x_2 = 2\sqrt{2}$, $y_2 = \frac{6-\sqrt{2}(2\sqrt{2})}{3} = \frac{6-4}{3} = \frac{2}{3}$. Point $B(2\sqrt{2}, \frac{2}{3})$.

Now, find the length of the chord $AB$:
$$AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$
$$AB = \sqrt{(2\sqrt{2}-0)^2 + \left(\frac{2}{3}-2\right)^2}$$
$$AB = \sqrt{8 + \left(-\frac{4}{3}\right)^2}$$
$$AB = \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{72+16}{9}} = \sqrt{\frac{88}{9}}$$
$$AB = \frac{\sqrt{4 \times 22}}{3} = \frac{2\sqrt{22}}{3}$$

Comparing with the given length $\frac{2\sqrt{\alpha}}{3}$:
$$\alpha = 22$$

Ans. (2)