I need help to find the parabola

Question

A parabola has focus (1,3) and directrix x=-5. Sketch the parabola. Identify the vertex of the parabola and the focal length c ( the distance from vertex to focus) and write the equation for the parabola?

Answer 1

F(1,3) D: x=-5

the vertex is half way between the focus and directrix

((-5+1)/2, 3) = (-2,3), V(-1,3)

It's a rightward opening parabola with general equation x=a(y-h)^2 + k, where (k,h) is the vertex = (-1,3)

x = a(y-3)^2 -2

x =a(y^2 -6y + 9) -2

x =ay^2 -6y +9a-2

9a-2 is the y intercept = 9(3/16)-2 = 27/16 -32/16 = -5/16 and 6 5/16 or (0,-5/16), (0, 6 5/16)

x intercept is only one: (-5/16, 0)

plug in any other point such as (1,7) on the parabola to solve for a. a>0 for rightward opening parabolas

(1,7) is equidistant from the directrix and focus, directly above the focus, to the right of the directrix

1 =a(7-3)^2 -2

3 =16a

a = 3/16

the parabola is

x = (3/16)(y-3)^2 -2