calculus
posted by Joseph .
let m and b be nonzero real numbers
if the line y=mx+b intersects y^2=4px in only one point, show that p=mb

calculus 
Reiny
sub the straight line into the parabola
(mx+b)^ = 4px
m^2x^2 + 2mbx + b^2  4px = 0
m^2x^2 + x(2mb4p) + b^2 = 0
to have one solution, (one intersection point), the discriminant has to be zero
(2mb4p)^2  4(m^2)(b^2) = 0
4 m^2b^2  16mbp + 16 p^2  4m^2b^2 = 0
16mbp  16p^2 = 0
16p(mb  p) = 0
so p = 0 or p = mb , but if p=0 we couldn't have a parabola, so
p = mb
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