parabolas and tangents?

Question

prove that a line tangent to the parabola x^2=4py at the point(a,b) crosses the y-axis at (0,-b)

Answer 1

Since the tangets touches the parabola at (a,b), then a^2 = 4pb, or 4p = a^2/b.

y = x^2/(a^2/b) = b*x^2/a^2.

y' = 2bx/a^2 so to find the slope of the tangent line at that point plug in the value: 2b(a)/a^2 = 2b/a.

Now you have a point and the slope, so use Point-Slope form of the line:

y - b = (2b/a)(x - a) --> (2b/a)x - 2b + b -->
y = (2b/a)x - b

The y-intercept of that line is (0,-b).

Answer 2

the slope of the tangent is
dy/dx = 2x/(4p) = x/(2p)
so the equation is
(y - b)/(x - a) = a/(2p)
y - b = a/(2p) (x - a)

at x = 0,
y = b - a^2/(2p),
but a^2 = 4pb, or b = a^2/(4p) or a^2/(2p) = 2b
so y = b - 2b = -b