JEE Mathematics 2025 Question with Solution

Given: A line through $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse

$$\frac{x^2}{36} + \frac{y^2}{25} = 1$$

at $A$ and $B$.

Find: The value of $5(PA^2 + PB^2)$ when $(PA)(PB)$ is maximum.

Assume any point on line $AB$ is

$$Q(\sqrt{5} + r\cos\theta, \sqrt{5} + r\sin\theta)$$

Substituting in the ellipse,

$$25(\sqrt{5} + r\cos\theta)^2 + 36(\sqrt{5} + r\sin\theta)^2 = 900$$

Expanding,

$$r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) + 25 \cdot 5 + 36 \cdot 5 = 900$$

So,

$$r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) - 595 = 0$$

If the roots in $r$ correspond to the directed distances to $A$ and $B$, then

$$PA \cdot PB = \frac{595}{25\cos^2\theta + 36\sin^2\theta}$$

To maximize $PA \cdot PB$, minimize the denominator:

$$25\cos^2\theta + 36\sin^2\theta = 25 + 11\sin^2\theta$$

This is minimum when $\sin^2\theta = 0$. Hence the line is parallel to the $x$-axis, so

$$y_A = y_B = \sqrt{5}$$

Putting $y = \sqrt{5}$ in the ellipse,

$$\frac{x^2}{36} + \frac{5}{25} = 1$$ $$\frac{x^2}{36} = \frac{4}{5}$$ $$x^2 = \frac{144}{5}$$

Thus,

$$A = \left(-\frac{12}{\sqrt{5}}, \sqrt{5}\right), \quad B = \left(\frac{12}{\sqrt{5}}, \sqrt{5}\right)$$

Now,

$$PA^2 + PB^2 = \left(\sqrt{5} - \frac{12}{\sqrt{5}}\right)^2 + \left(\sqrt{5} + \frac{12}{\sqrt{5}}\right)^2$$ $$= 2\left(5 + \frac{144}{5}\right) = \frac{338}{5}$$

Therefore,

$$5(PA^2 + PB^2) = 338$$

The correct option is D.

Given: The ellipse is

$$\frac{x^2}{36} + \frac{y^2}{25} = 1$$

and the line passes through $P(\sqrt{5}, \sqrt{5})$.

Find: $5(PA^2 + PB^2)$ when $(PA)(PB)$ is maximum.

Use the parametric form of the line:

$$x = \sqrt{5} + r\cos\theta, \quad y = \sqrt{5} + r\sin\theta$$

Substitute into the ellipse:

$$\frac{(\sqrt{5} + r\cos\theta)^2}{36} + \frac{(\sqrt{5} + r\sin\theta)^2}{25} = 1$$

Multiplying by $900$,

$$25(\sqrt{5} + r\cos\theta)^2 + 36(\sqrt{5} + r\sin\theta)^2 = 900$$

Expand:

$$25(5 + 2\sqrt{5}r\cos\theta + r^2\cos^2\theta) + 36(5 + 2\sqrt{5}r\sin\theta + r^2\sin^2\theta) = 900$$ $$(125 + 50\sqrt{5}r\cos\theta + 25r^2\cos^2\theta) + (180 + 72\sqrt{5}r\sin\theta + 36r^2\sin^2\theta) = 900$$

Hence,

$$r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) - 595 = 0$$

The product of distances is

$$PA \cdot PB = \frac{595}{25\cos^2\theta + 36\sin^2\theta}$$

Now,

$$25\cos^2\theta + 36\sin^2\theta = 25(1 - \sin^2\theta) + 36\sin^2\theta = 25 + 11\sin^2\theta$$

This is smallest when $\sin^2\theta = 0$, so the required line is horizontal:

$$y = \sqrt{5}$$

Substitute in the ellipse:

$$\frac{x^2}{36} + \frac{5}{25} = 1 \implies \frac{x^2}{36} = \frac{4}{5}$$ $$x = \pm \frac{12}{\sqrt{5}} = \pm \frac{12\sqrt{5}}{5}$$

So the intersection points are

$$A\left(\frac{12\sqrt{5}}{5}, \sqrt{5}\right), \quad B\left(-\frac{12\sqrt{5}}{5}, \sqrt{5}\right)$$

Therefore,

$$PA = \left|\frac{12\sqrt{5}}{5} - \sqrt{5}\right| = \frac{7\sqrt{5}}{5}$$ $$PB = \left|-\frac{12\sqrt{5}}{5} - \sqrt{5}\right| = \frac{17\sqrt{5}}{5}$$

Then,

$$PA^2 = \left(\frac{7\sqrt{5}}{5}\right)^2 = \frac{49}{5}$$ $$PB^2 = \left(\frac{17\sqrt{5}}{5}\right)^2 = \frac{289}{5}$$ $$PA^2 + PB^2 = \frac{49}{5} + \frac{289}{5} = \frac{338}{5}$$

Thus,

$$5(PA^2 + PB^2) = 338$$

Therefore, the correct option is D.