Prove parabolas intersect at right angles (polar eq'ns)

[subwaybusker]

Homework Statement

Show that the parabolas r=c/(1+cosθ)and r'=d/(1-cosθ) intersect at right angles.

The Attempt at a Solution

I found the points of intersection by setting the two equations equal, to which I got:
cosθ = (c- d)/(c+d)
θ = cos^-1[(c- d)/(c+d)]

then i tried to find the slope of the two equations:

x=dcosθ/1-cosθ ; y=dsinθ/1-cosθ

dy/dθ = [dcosθ(1-cosθ)-(sinθ)dsinθ] / (1-cosθ)^2 = d(cosθ-1)/(1-cosθ)^2
dx/dθ= [-dsinθ(1-cosθ)-(sinθ)dcosθ] / (1-cosθ)^2 = -dsinθ/(1-cosθ)^2

dy/dx=cosθ-1/-sinθ

x=ccosθ/1+cosθ ; y=csinθ/1+cosθ

dy'/dθ = [-csinθ(1+cosθ)-(-sinθ)ccosθ] / (1+cosθ)^2 = -csinθ/(1+cosθ)^2
dx'/dθ= [ccosθ(1+cosθ)-(-sinθ)csinθ] / (1+cosθ)^2 = c(cosθ+1)/(1+cosθ)^2

dy/dx=cosθ+1/-sinθ

Then I don't know what to do

 

[Dick]

Two slopes are perpendicular if their product is -1. Are they? Use a trig identity.

 

[subwaybusker]

thanks!

 

[Dick]

You don't have to assume that. sin(x)^2+cos(x)^2=1. So cos(x)^2-1=-sin(x)^2. ALWAYS. It's an identity.