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

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(A) The parabola is $y^2=4x$,so its focus $S$ is $(1, 0)$.Let $P$ be $(t^2, 2t)$. The slope of the focal chord $PS$ is given by $\frac{2t-0}{t^2-1} =

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$15$

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(A) The parabola is $y^2=4x$,so its focus $S$ is $(1, 0)$.Let $P$ be $(t^2, 2t)$. The slope of the focal chord $PS$ is given by $\frac{2t-0}{t^2-1} = 	an 60^{\circ} = \sqrt{3}$.$2t = \sqrt{3}(t^2-1) \Rightarrow \sqrt{3}t^2 - 2t - \sqrt{3} = 0$.Solving for $t$,we get $t = \frac{2 \pm \sqrt{4 - 4(\sqrt{3})(-\sqrt{3})}}{2\sqrt{3}} = \frac{2 \pm 4}{2\sqrt{3}}$.Since $P$ is in the first quadrant,$t > 0$,so $t = \frac{6}{2\sqrt{3}} = \sqrt{3}$.Thus,$P = ((\sqrt{3})^2, 2\sqrt{3}) = (3, 2\sqrt{3})$.The circle with diameter $PS$ has the equation $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$,where $P=(3, 2\sqrt{3})$ and $S=(1, 0)$.$(x-3)(x-1) + (y-2\sqrt{3})(y-0) = 0$.$x^2 - 4x + 3 + y^2 - 2\sqrt{3}y = 0$.Since the circle touches the $y$-axis at $(0, \alpha)$,we set $x=0$:$3 + y^2 - 2\sqrt{3}y = 0$.This is a quadratic in $y$ with a single solution (since it touches),so the discriminant is $0$,or simply $y^2 - 2\sqrt{3}y + 3 = 0 \Rightarrow (y-\sqrt{3})^2 = 0$.Thus,$\alpha = \sqrt{3}$.Then $5\alpha^2 = 5(\sqrt{3})^2 = 5 	imes 3 = 15$.

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