Analytic Geometry – Problem 13: find

For $C:x^2=4y$, line $l:y=kx+4$ meets at $A,B$. If $P\in C$ and $\angle PFA=\angle PFB$, find $m_{PF}m_l$.

Let $A(x_1,y_1),B(x_2,y_2)$. Intersect $$y=kx+4$$ with $$x^2=4y$$ to get $$x^2-4kx-16=0,$$ hence $$x_1+x_2=4k,\qquad x_1x_2=-16.$$

Let $F(0,1)$, $P(x_0,y_0)$. The angle condition $\angle PFA=\angle PFB$ is the internal-bisector condition in vector form: $$\frac{\overrightarrow{FA}\cdot\overrightarrow{FP}}{|FA|}=\frac{\overrightarrow{FB}\cdot\overrightarrow{FP}}{|FB|}.$$ For parabola $x^2=4y$, distance to focus equals distance to directrix $y=-1$, so $$|FA|=y_1+1,\qquad |FB|=y_2+1.$$ Substitute $A,B,P$ and use $y_i=kx_i+4$ together with $x_1+x_2=4k, x_1x_2=-16$; after cancellation we obtain $$\frac{y_0-1}{x_0}=-\frac{5}{2k}.$$ Now $k_{PF}=\dfrac{y_0-1}{x_0}$, $m_l=k$, therefore $$k_{PF}m_l=\left(-\frac{5}{2k}\right)k=-\frac52.$$

$$m_{PF}m_l=-\frac52.$$