Let the focal chord $P Q$ 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…

Question

Let the focal chord $P Q$ 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 :

Aicharya · Accepted Answer

$	an 60^{\circ}=\frac{2 \mathrm{t}-0}{\mathrm{t}^2-1}=\sqrt{3} \Rightarrow \mathrm{t}=\sqrt{3}$ $	herefore \mathrm{P}(3,2 \sqrt{3})$ Circle : $(x-1)(x-3)+(y-0)(y-2 \sqrt{3})=0$ at $x=0$ $\Rightarrow 3+y^2-2 \sqrt{3} y=0$ $\Rightarrow \mathrm{y}=\sqrt{3}=\alpha$ $5 \alpha^2=15$ .

Let the focal chord $P Q$ of the parabola $y^2=4 x$ make an angle of $60^{\circ}$ with the positive $x$-axis…

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