Maths Q9 – JEE-MAIN 2025

The axis of a parabola is the line

y = x

and its vertex and focus are in the first quadrant at distances

\sqrt{2}

and

2 \sqrt{2}

units from the origin, respectively. If the point

(1, k)

lies on the parabola, then a possible value of

k

is :

Clue 1 of 3: Geometric Properties of Parabola

Understand the relationship between the vertex, focus, axis, and directrix of a parabola, especially when the axis is not parallel to the coordinate axes.

🧠 Full Solution Path

Step 1: Determine Vertex and Focus Coordinates✦ Active

The axis of the parabola is $y = x$ . Since the vertex $V$ and focus $F$ are in the first quadrant and on the axis, their coordinates are of the form $(x_{0}, x_{0})$ . The distance of $V$ from the origin is $\sqrt{2}$ , so $\sqrt{x_{V}^{2} + x_{V}^{2}} = \sqrt{2} ⟹ \sqrt{2 x_{V}^{2}} = \sqrt{2} ⟹ x_{V} \sqrt{2} = \sqrt{2} ⟹ x_{V} = 1$ . Thus, the vertex is $V = (1, 1)$ . The distance of $F$ from the origin is $2 \sqrt{2}$ , so $\sqrt{x_{F}^{2} + x_{F}^{2}} = 2 \sqrt{2} ⟹ \sqrt{2 x_{F}^{2}} = 2 \sqrt{2} ⟹ x_{F} \sqrt{2} = 2 \sqrt{2} ⟹ x_{F} = 2$ . Thus, the focus is $F = (2, 2)$ . The distance between the vertex and focus is $a = V F = \sqrt{(2 - 1)^{2} + (2 - 1)^{2}} = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$ .

Step 2: Find the Equation of the Directrix○ Expand

The directrix is perpendicular to the axis $y = x$ (slope $1$ ) and passes through a point $D$ such that $V$ is the midpoint of $F D$ . The slope of the directrix is $- 1$ . Let $D = (x_{D}, y_{D})$ . Since $V = (1, 1)$ is the midpoint of $F (2, 2)$ and $D (x_{D}, y_{D})$ , we have $1 = \frac{2 + x_{D}}{2}$ and $1 = \frac{2 + y_{D}}{2}$ . This implies $x_{D} = 0$ and $y_{D} = 0$ . So, the directrix passes through the origin $(0, 0)$ . The equation of the directrix is $y - 0 = - 1 (x - 0) ⟹ y = - x ⟹ x + y = 0$ .

Step 3: Formulate Parabola Equation and Solve for k○ Expand

By the definition of a parabola, for any point $P (x, y)$ on the parabola, its distance from the focus $F (2, 2)$ is equal to its distance from the directrix $x + y = 0$ . So, $P F^{2} = P D^{2}$ .

(x - 2)^{2} + (y - 2)^{2} = {(\frac{x + y}{\sqrt{1^{2} + 1^{2}}})}^{2}

Simplifying the equation:

2 [(x - 2)^{2} + (y - 2)^{2}] = (x + y)^{2}

⟹ 2 (x^{2} - 4 x + 4 + y^{2} - 4 y + 4) = x^{2} + 2 x y + y^{2}

⟹ 2 x^{2} - 8 x + 8 + 2 y^{2} - 8 y + 8 = x^{2} + 2 x y + y^{2}

⟹ x^{2} + y^{2} - 2 x y - 8 x - 8 y + 16 = 0

The point $(1, k)$ lies on the parabola. Substitute $x = 1$ and $y = k$ into the equation:

1^{2} + k^{2} - 2 (1) k - 8 (1) - 8 k + 16 = 0

⟹ 1 + k^{2} - 2 k - 8 - 8 k + 16 = 0

⟹ k^{2} - 10 k + 9 = 0

Factoring the quadratic equation:

(k - 1) (k - 9) = 0

The possible values for $k$ are $1$ or $9$ . Among the given options, $9$ is a possible value.

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Maths Question 9 – JEE-MAIN 2025

Clue 1 of 3: Geometric Properties of Parabola

🧠 Full Solution Path