Posted by **Joseph ** on Tuesday, May 24, 2011 at 8:13pm.

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**, Tuesday, May 24, 2011 at 8:51pm
sub the straight line into the parabola

(mx+b)^ = 4px

m^2x^2 + 2mbx + b^2 - 4px = 0

m^2x^2 + x(2mb-4p) + b^2 = 0

to have one solution, (one intersection point), the discriminant has to be zero

(2mb-4p)^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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