Common Normal Parabola Problem

Question

Prove that two parabolas $y^2=4ax $ and $y^2=4c(x-b)$ cannot have a common normal other than the axis, unless $ b/a-c>2$. I couldn't think of a satisfactory approach. Please Help.

Answer 1

Consider a point $P_1\equiv (x_1,y_1)$ on the first parabola and a point $P_2\equiv (x_2,y_2)$ on the second one with $$x_1=\frac{y_1^2}{4a}$$and$$x_2=\frac{y_2^2}{4c}+b$$ Implicitly differentiate both equations of the parabolas with respect to $x$ to obtain the slopes of the tangents.
Then (1) equate the slopes of the normals and (2) put the slope of $P_1P_2$ equal to the common slope of the normals. Couple both conditions (1) and (2) to get the system $$\begin {cases} -\frac {y_1}{2a}=-\frac {y_2}{2c} \\ \\ \frac {y_1-y_2}{x_1-x_2}=-\frac {y_1}{2a} \end{cases} $$ You find $$y_1=0$$ and $$y_1^2=4a^2 \cdot \frac {b-2(a-c)}{a-c}$$ hence the reality condition is $$\frac b{a-c}>2$$