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

Question

Let the focal chord $PQ$ of the parabola $y^2=4x$ 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

The tangent at $P=(t^2,2t)$ is $ty=x+t^2$, with slope $1/t$. Since the circle on diameter $PS$ touches the $y$-axis, the geometry gives the tangent-line slope equal to $\tan30^\circ$ (half the $60^\circ$ chord angle):

$$\frac1t=\tan30^\circ=\frac{1}{\sqrt3} \;\Rightarrow\; t=\sqrt3.$$

The point of tangency has ordinate $\alpha=at=\sqrt3$ (with $a=1$), so $5\alpha^2=5\times3=15$.

Solutions · 1