Question

How do you find the standard equation given focus (8,10), and vertex (8,6)?

Answer 1

The focus is above the vertex, therefore, the vertex form of the equation is:

`y = a(x-h)^2+k`

Use the focus to compute `a = 1/(4f)`

Expand the equation into standard form.

Explanation:

The focal distance, f, is the distance form the vertex to the focus:

`f = 10-6`

`f = 4`

Compute the value of "a":

`a = 1/(4f)`

`a = 1/16`

The vertex tells us that `h = 8 and k = 6`

Substituting these values into the vertex form:

`y = 1/16(x-8)^2+6`

Expand the square:

`y = 1/16(x^2-16x+64)+6`

`y = x^2/16-x+4+6`

`y = x^2/16-x+10` This is standard form.

Answer 2

Find the distance between vertex and focus. Call that p.
Since it opens upward, p >0. Use (x-h)2 = 4p(y - k).

Explanation:

The equation of the parabola that opens up or down and has vertex (h, k) is
`(x-h)^2 = 4p(y - k)`
where p is the difference between the y-coordinates of the focus and the vertex.

In this example, p = 10 - 6 = 4, and (h, k) = (8, 6). Therefore,

`(x-8)^2 = 16(y - 6)`

This may have been the form you were seeking.
[If we want this in the "standard form," that usually means solving for the variable that is not squared. Distribute the 16, and add...

`(x-8)^2 = 16(y - 6)`
`(x-8)^2 = 16y - 96`
`(x-8)^2 + 96 = 16y`
`y = 1/16 (x-8)^2 + 6`
]
If you were only interested in the standard form, set a = 1/(4p) and go straight to the vertex form:

a = 1/(4*4) = 1/16. Therefore,

`y = a(x-h)^2 + k`
That is,
`y = 1/16 (x-8)^2 + 6`

Use FOIL and distribute if you prefer the form
`y = ax^2 + bx + c`

I will not spoil that fun for you.