JEE Mathematics 2025 Question with Solution

Given: The axis of the parabola is $y = x$. Its vertex and focus are in the first quadrant at distances $\sqrt{2}$ and $2\sqrt{2}$ from the origin, respectively. The point $(1, k)$ lies on the parabola.

Find: A possible value of $k$.

From the solution, take the vertex as $(0,0)$, the focus as $(2\sqrt{2}, 2\sqrt{2})$, and the directrix as $x + y = 0$.

Using the definition of a parabola, the distance from any point on the parabola to the focus equals the distance from that point to the directrix.

Let $P(1,k)$ be a point on the parabola.

Distance from $P$ to the focus:

$$PS = \sqrt{(1 - 2\sqrt{2})^2 + (k - 2\sqrt{2})^2}$$

Distance from $P$ to the directrix $x + y = 0$:

$$PM = \frac{|1 + k|}{\sqrt{2}}$$

Since $P$ lies on the parabola,

$$\sqrt{(1 - 2\sqrt{2})^2 + (k - 2\sqrt{2})^2} = \frac{|1 + k|}{\sqrt{2}}$$

After solving this equation, we get $k = 9$.

Therefore, a possible value of $k$ is $9$, so the correct option is B.

Given: The directrix is $x + y = 0$ and the condition is $PS = PM$, where $P$ is a point on the parabola.

Find: The value of $K$.

The second approach in the solution writes the working as:

$$\sqrt{(1 - 2)^2 + (K - 2)^2} = \frac{(1 + K)}{\sqrt{2}}$$

Then it simplifies to:

$$2K^2 + 8K - 8K + 2 = K^2 + 1 + 2K$$

So,

$$K^2 - 10K + 9 = 0$$

Solving the quadratic equation gives:

$$K = 9$$

Therefore, the correct option is B.