Find the standard form of the equation of the parabola with the given characteristics. Focus (5, 1); Directrix: x=-3

Answer 1

Focus (5,1) Directrix x=-3

means a rightward opening parabola with vertex x coordinarte = 1 (the average of 5 and -3)

the axis of symmetry is y=1. Vertex = (1,1) The vertex is midway between the Focus and Directrix

x = a(y-1)^2 +1

to solve for the coefficient "a," plug in another point on the parabola, directly above the Focus

it helps to draw a rough graph. The point (5,9) is 8 units vertically from the Focus and 8 units horizontally from the Directrix.

5=a(9-1)^2 + 1

5 =64a + 1

64a = 4

a = 4/64 = 1/16

x= (1/16)(y-1)^2 + 1 is in vertex form

x = (1/16)(y^2 -2y + 1) +1

x = (1/16)y^2 -(1/8)y + 17/16 is in standard form,

x = y^2/16 - y/8 + 17/16

x =ay^2 + by +c where a=1/16, b = -1/8 and c= 17/16 = x intercept. There is no y intercept