Maths Q1 – JEE-MAIN 2025

Let the focal chord PQ of the parabola

y^{2} = 4 x

make an angle of

60^{\circ}

with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the y-axis at the point

(0, α)

, then

5 α^{2}

is equal to :

Clue 1 of 3: Parabola and Focal Chord Properties

First, identify the focus of the given parabola and then use the equation of a line to find the coordinates of point P on the parabola.

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● Video Walkthrough

Step 1: Determine Parabola Properties and Point P✦ Active

The parabola is $y^{2} = 4 x$ , so comparing with $y^{2} = 4 a x$ , we get $a = 1$ . The focus is $S = (a, 0) = (1, 0)$ . The focal chord PQ makes an angle of $60^{\circ}$ with the x-axis, so its slope is $m = \tan (60^{\circ}) = \sqrt{3}$ . The equation of the focal chord passing through $S (1, 0)$ is $y - 0 = \sqrt{3} (x - 1)$ , or $y = \sqrt{3} (x - 1)$ . Substitute this into the parabola equation: $(\sqrt{3} (x - 1))^{2} = 4 x ⟹ 3 (x - 1)^{2} = 4 x ⟹ 3 (x^{2} - 2 x + 1) = 4 x ⟹ 3 x^{2} - 6 x + 3 = 4 x ⟹ 3 x^{2} - 10 x + 3 = 0$ . Factoring the quadratic equation gives $(3 x - 1) (x - 3) = 0$ , so $x = 3$ or $x = \frac{1}{3}$ . Since P lies in the first quadrant ( $x_{P} > 0, y_{P} > 0$ ), for $x = 3$ , $y = \sqrt{3} (3 - 1) = 2 \sqrt{3}$ . Thus, $P = (3, 2 \sqrt{3})$ . (For $x = 1 / 3$ , $y = - 2 \sqrt{3} / 3$ , which is in the fourth quadrant, so it's point Q).

Step 2: Find Circle Equation○ Expand

The circle has PS as its diameter, where $P = (3, 2 \sqrt{3})$ and $S = (1, 0)$ . The center of the circle $C$ is the midpoint of PS: $C = (\frac{3 + 1}{2}, \frac{2 \sqrt{3} + 0}{2}) = (2, \sqrt{3})$ . The radius squared $r^{2}$ is ${(\frac{P S}{2})}^{2} = \frac{(3 - 1)^{2} + (2 \sqrt{3} - 0)^{2}}{4} = \frac{2^{2} + (2 \sqrt{3})^{2}}{4} = \frac{4 + 12}{4} = \frac{16}{4} = 4$ . So, the radius $r = 2$ . The equation of the circle is $(x - 2)^{2} + (y - \sqrt{3})^{2} = 4$ .

Step 3: Determine

α

and Calculate

5 α^{2}

○ Expand

The circle touches the y-axis at the point $(0, α)$ . For a circle with center $(h, k)$ and radius $r$ to touch the y-axis, the absolute value of its x-coordinate of the center must be equal to its radius, i.e., $| h | = r$ . Here, $| 2 | = 2$ , which matches the radius $r = 2$ . The point of tangency on the y-axis is $(0, k)$ , where $k$ is the y-coordinate of the center. Therefore, $(0, α) = (0, \sqrt{3})$ , which implies $α = \sqrt{3}$ . Finally, we need to calculate $5 α^{2} = 5 (\sqrt{3})^{2} = 5 \times 3 = 15$ .

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Maths Question 1 – JEE-MAIN 2025

Clue 1 of 3: Parabola and Focal Chord Properties

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