Find the standard form of the equation of the parabola with the given characteristics. Vertex: (6, 1); focus: (4, 1)

Answer 1

It's a parabola opening to the left

x-6=-a(y-1)^2

axis of symmetry is y=1

directrix line is x=8

find a point directly above the focus on the parabola. It's (4,5) as it must have an equal distance to the focus as to the directrix. It's a distance of 4 to either.

Now plug that point into the parabola equation to get "a"

4-6 =-a(5-1)^2

-2 = -16a

a = 1/8

the parabola equation is

x-6 =-(1/8)(y-1)^2, expand if you want, maybe multiply by 8 to eliminate the fraction

8x -48 = -y^2 + 2y -1

8x = -y^2 + 2y +47 (that gives you the y intercept as 47/8 or the point (0,47/8=almost 6))