Angle between the parabolas $y^2 = 4b (x – 2a + b)$ and $x^2 + 4a (y – 2b – a) = 0$ at the common end of their latus rectum is?

Question

Angle between the parabolas $y^2 = 4b (x – 2a + b)$ and $x^2 + 4a (y – 2b – a) = 0$ at the common end of their latus rectum is ?

Solving the two equations to find the point of intersection seems terribly long. Is there any easier method ?

Answer 1

You don't need to solve the system.

The equation of the directrix of $(y-n)^2=4p(x-m),(x-m)^2=4p(y-n)$ is $x=m-p,y=n-p$ respectively.

Using this fact gives that the equation of the directrix of $y^2=4b(x-2a+b),x^2=-4a(y-2b-a)$ is $x=2a-2b,y=2a+2b$ respectively. So, the latus rectum is on the line $x=2a,y=2b$ respectively.

Hence, the coordinates of the common end of their latus rectum is $(2a,2b)$.

Now $$y^2=4b(x-2a+b)\implies 2yy'=4b\implies y'=\frac{2b}{y}$$ $$x^2=-4a(y-2b-a)\implies 2x=-4ay'\implies y'=-\frac{x}{2a}$$ Since we want to find the angle between $$y-2b=\frac{2b}{2b}(x-2a)$$ and $$y-2b=-\frac{2a}{2a}(x-2a)$$ we get that the answer is $\color{red}{\pi/2}$.