Find, in standard form, the equation of : the parabola with focus at (3,5) and directrix at x=-1

Answer 1

I don't like to show you by just plugging into some formula, since that way you really don't learn much
So I will do it by using the actual definition of the parabola

let P(x,y) be any point on our parabola.
P must be equidistant from (3,5) and the line x = -1

so x - (-1) = √( (x-3)^2 + (y-5)^2)
square both sides, and expanding
x^2 + 2x + 1 = x^2 - 6x + 9 + y^2 - 10y + 25 **
8x = y^2 - 10y + 33
x = (1/8)(y^2 - 10y + 33)

suppose we had left ** as
x^2 + 2x + 1 = x^2 - 6x + 9 + (y-5)^2
8x = (y-5)^2 + 8
x = (1/8)(y-5)^2 + 1

check my arithmetic

you might also look at this video by the Khan Academy
The parabola has a vertical directrix, yours has a horizontal directrix, but he explains it well
https://www.khanacademy.org/math/algebra2/intro-to-conics-alg2/focus-and-directrix-of-a-parabola-alg2/v/equation-for-parabola-from-focus-and-directrix

Answer 2

Recall that the parabola

y^2 = 4px

has vertex at (0,0) and the distance from focus to directrix is 2p.

Your parabola's distance from focus to directrix is 4, so p=2. The focus is at (3,5), so the vertex halfway between the focus and directrix, at (1,5). So, your parabola is

(y-5)^2 = 8(x-1)

which agrees with Reiny's result.