# How do you find vertices, foci, directrix, axis of symmetry of Y+2 =1/10(x-6)^2?

Nov 29, 2016

The vertex is $\left(6 , - 2\right)$
The focus is $= \left(6 , \frac{1}{2}\right)$
The directrix is $y = - \frac{9}{2}$
The axis of symmetry is $x = 6$

#### Explanation:

We rewrite the equation as

${\left(x - 6\right)}^{2} = 10 \left(y + 2\right)$

We compare this to the equation

${\left(x - a\right)}^{2} = 2 p \left(y - b\right)$

The vertex is $\left(a , b\right) = \left(6 , - 2\right)$

$2 p = 10$, $\implies$, $p = 5$

The focus is $\left(a , b + \frac{p}{2}\right) = \left(6 , - 2 + \frac{5}{2}\right) = \left(6 , \frac{1}{2}\right)$

The directrix is $y = b - \frac{p}{2} = - 2 - \frac{5}{2} = - \frac{9}{2}$

The axis of symmetry is $x = 6$
graph{(x-6)^2=10(y+2) [-36.54, 36.54, -18.27, 18.28]}