Two straight line are at right angles to one another, one of them touches the parabola $y^2=4a(x+a)$ and the other touches the parabola $y^2=4a'(x+a')$. Show that the point of intersection of the straight lines will lie on the straight line $x+a+a'=0$.
I tried to solve this problem, I find two tangents of two parabolas, then they meet at right angle,$(m_1.m_2=-1)$. But I can not find the result. Someone please help.
Let $y=mx+c$ be tangent to the curve $y^2=4a(x+a)$
Solving simultaneously gives the quadratic $$m^2x^2+x(2mc-4a)+c^2-4a^2=0$$
This has zero discriminant as a condition for tangency, so after a couple of lines of algebra, we get $$\frac{m}{1+m^ 2}=\frac ac$$
Now let the perpendicular line have equation $y=-\frac 1m x+c'$ which is tangent to the curve $y^2=4a'(x+a')$
In which case, without further calculation, we have $$\frac{-\frac 1m}{1+\frac{1}{m^2}}=\frac{a'}{c'}$$
Therefore, $$\frac{m}{1+m^ 2}=\frac ac=-\frac {a'}{c'}$$
Meanwhile, solving the two lines simultaneously gives $$x=\frac{m}{m^2+1}(c'-c)$$
So, using the above results, the locus is $$x=-a'-a$$ as required.