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

Mar 25, 2017

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

$y = a {\left(x - h\right)}^{2} + k$

Use the focus to compute $a = \frac{1}{4 f}$

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":

$a = \frac{1}{4 f}$

$a = \frac{1}{16}$

The vertex tells us that $h = 8 \mathmr{and} k = 6$

Substituting these values into the vertex form:

$y = \frac{1}{16} {\left(x - 8\right)}^{2} + 6$

Expand the square:

$y = \frac{1}{16} \left({x}^{2} - 16 x + 64\right) + 6$

$y = {x}^{2} / 16 - x + 4 + 6$

$y = {x}^{2} / 16 - x + 10$ 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
${\left(x - h\right)}^{2} = 4 p \left(y - k\right)$
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,

${\left(x - 8\right)}^{2} = 16 \left(y - 6\right)$

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...

${\left(x - 8\right)}^{2} = 16 \left(y - 6\right)$
${\left(x - 8\right)}^{2} = 16 y - 96$
${\left(x - 8\right)}^{2} + 96 = 16 y$
$y = \frac{1}{16} {\left(x - 8\right)}^{2} + 6$
]
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,

$y = a {\left(x - h\right)}^{2} + k$
That is,
$y = \frac{1}{16} {\left(x - 8\right)}^{2} + 6$

Use FOIL and distribute if you prefer the form
$y = a {x}^{2} + b x + c$

I will not spoil that fun for you.