JEE MAIN - Mathematics (2025 - 2nd April Morning Shift - No. 6)
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 $(0, \alpha)$, then $5 \alpha^2$ is equal to:
15
25
20
30
Explanation
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$$\begin{aligned} & P T: t y=x+a t^2 \\ & P S=P T \\ & M_t=\frac{1}{t}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ & t=\sqrt{3} \\ & \alpha=a t=\sqrt{3} \quad(a=1) \\ & \therefore 5 \alpha^2=15 \end{aligned}$$
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