find the equation of the parabola with vertex on  Oy, axis parallel to Ox, and passing through (2,2),(8,-1). conic sections the parabola. analytic geometry. how to do this? please help

Expert Answers

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Here's the solution.

The vertex lies on Oy, let the vertex of the parabola be (0, y0). The axis is parallel to Ox, this gives a parabola of the form (y - y0)^2 = 4ax

As it passes through (2,2) and (8,-1)

(2 - y0)^2 = 8a and (-1 - y0)^2 = 32a

=> (-1 - y0)^2 = 4*(2 - y0)^2

=> (-1 - y0 - 4 + 2y0)(-1 - y0 + 4 - 2y0) = 0

=> y0 = 5 and y0 = 1

a = (2 - 5)^2 / 8 = 9/8 and a = 1/8

The required parabola can have two equations:

(y - 5)^2 = 9x/2 and (y - 1)^2 = x/2

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