If P is a point on the parabola y^{2}=4ax wit | vedclass

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(A) The equation of the parabola is $y^{2}=4ax$. Let the coordinates of point $P$ on the parabola be $(at^{2}, 2at)$. The focus $F$ is $(a, 0)$ and th

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(A) The equation of the parabola is $y^{2}=4ax$. Let the coordinates of point $P$ on the parabola be $(at^{2}, 2at)$. The focus $F$ is $(a, 0)$ and the directrix is $x = -a$. The foot of the perpendicular $Q$ from $P(at^{2}, 2at)$ onto the directrix $x = -a$ is $(-a, 2at)$. By the definition of a parabola, the distance $PF$ is equal to the distance $PQ$. $PQ = \sqrt{(at^{2} - (-a))^{2} + (2at - 2at)^{2}} = \sqrt{(a(t^{2}+1))^{2}} = a(t^{2}+1)$. $PF = \sqrt{(at^{2}-a)^{2} + (2at-0)^{2}} = \sqrt{a^{2}(t^{2}-1)^{2} + 4a^{2}t^{2}} = \sqrt{a^{2}(t^{4}-2t^{2}+1+4t^{2})} = \sqrt{a^{2}(t^{2}+1)^{2}} = a(t^{2}+1)$. Since $PQ = PF$, the triangle $	riangle PQF$ is an isosceles triangle with $PQ = PF$. In an isosceles triangle, the angles opposite to the equal sides are equal, so $\angle PQF = \angle PFQ$. Therefore, $	an \angle PQF = 	an \angle PFQ$. Hence, $\frac{	an \angle PQF}{	an \angle PFQ} = 1$.

If $P$ is a point on the parabola $y^{2}=4ax$ with focus $F$. Let $Q$ denote the foot of the perpendicular from $P$ onto the directrix. Then, $\frac{\tan \angle PQF}{\tan \angle PFQ}$ is

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