Let the focal chord PQ of the parabola y^2=4 x make an angle of 60^ with the positive x ax…

Question

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:

Accepted Answer

$$\begin{aligned}
& P T: t y=x+a t^2 \
& P S=P T \
& M_t=\frac{1}{t}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \
& t=\sqrt{3} \
& \alpha=a t=\sqrt{3} \quad(a=1) \
& \therefore 5 \alpha^2=15
\end{aligned}$$

Solutions · 1