Assuming k is nonzero...
kx^2=k(x-2)^2
iff
kx^2-k(x-2)^2=0
factor
k(x^2-(x-2)^2)=0
iff
x^2-(x-2)^2=0, since k is non zero.
iff x^2-(x^2-4x+4)=0
x^2-x^2+4x-4=0
4x=4
x=1, not 2.
Hey guys, I have a question that I am stuck atm and need some help with.
"For what values of the nonzero constant k do the curves y=kx^2 and y=k(x-2)^2 intersect at right angles?"
I started off by finding the intersection point of the 2 curves:
kx^2 = k(x-2)^2
x = 2
After which I differentiate to find my gradient or slope:
For the curve y = kx^2, m1 = 2kx
For the curve y = k(x-2)^2, m2 = 2k(x-2)
I understand that since the 2 curves intersect at 90 degrees, m1.m2 = -1
But when i sub x=2 into my m2, i get 0 which gives me m1.0 = -1
I am definitely wrong somewhere but I cannot really figure it out. Am I even on the right track to solving this question?
Cheers,
Dickson