JEE Mathematics 2025 Question with Solution

Given: The parabola is $y^2 = 16x$ and point $P(1,-4)$ lies on the focal chord $PQ$.

Find: If the focus divides $PQ$ in the ratio $m:n$, find $m^2+n^2$.

Compare with the standard form

$$y^2 = 4ax$$

So,

$$4a = 16 \implies a = 4$$

Hence the focus is

$$S = (4,0)$$

A general point on the parabola is

$$(at^2, 2at)$$

For point $P(1,-4)$,

$$2at_1 = -4$$

Substituting $a=4$,

$$8t_1 = -4 \implies t_1 = -\frac{1}{2}$$

Since $PQ$ is a focal chord, the parameters satisfy

$$$$

Therefore,

$$\left(-\frac{1}{2}\right)t_2 = -1 \implies t_2 = 2$$

So the other end point is

$$Q(at_2^2, 2at_2) = (4\cdot 2^2, 2\cdot 4\cdot 2) = (16,16)$$

Now compute the distances from the focus:

$$PF = \sqrt{(4-1)^2 + (0-(-4))^2} = \sqrt{3^2+4^2} = 5$$ $$FQ = \sqrt{(16-4)^2 + (16-0)^2} = \sqrt{12^2+16^2} = 20$$

Hence,

$$\frac{m}{n} = \frac{PF}{FQ} = \frac{5}{20} = \frac{1}{4}$$

So $m=1$ and $n=4$. Therefore,

$$m^2+n^2 = 1^2+4^2 = 1+16 = 17$$

Thus, the correct option is A.

Given: The parabola is $y^2 = 16x$, point $P(1,-4)$ is one end of a focal chord, and the focus is required to divide chord $PQ$ in the ratio $m:n$.

Find: The value of $m^2+n^2$.

From

$$y^2 = 16x = 4ax$$

we get

$$a=4$$

and the focus is

$$(4,0)$$

Write point $P$ in parametric form:

$$P=(at_1^2,2at_1)$$

Using $P=(1,-4)$,

$$8t_1=-4 \implies t_1=-\frac{1}{2}$$

For a focal chord,

$$t_1t_2=-1$$

Hence,

$$t_2=2$$

and so

$$Q=(at_2^2,2at_2)=(16,16)$$

Let the focus divide $PQ$ internally in the ratio $\lambda:1$. Using the section formula for the $y$-coordinate,

$$\frac{\lambda \cdot 16 + 1\cdot(-4)}{\lambda+1} = 0$$

Thus,

$$16\lambda - 4 = 0 \implies \lambda = \frac{1}{4}$$

Therefore,

$$\frac{m}{n} = \frac{1}{4}$$

So,

$$m^2+n^2 = 1^2+4^2 = 17$$

Therefore, the required value is $17$ and the correct option is A.