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

2 Answers
Mar 25, 2017

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.

Mar 25, 2017

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.