# What is the equation for a parabola with a vertex at (5,-1) and a focus at (3,-1)?

Mar 25, 2017

$x = - \frac{1}{8} {\left(y + 1\right)}^{2} + 5$

#### Explanation:

Since the $y$-coordinates of the vertex and focus are the same, vertex is to the right of focus.

Hence, this is a regular horizontal parabola and as vertex $\left(5 , - 1\right)$ is to the right of focus, it opens to the left.and $y$ part is squared.

Therefore, equation is of the type

${\left(y + 1\right)}^{2} = - 4 p \left(x - 5\right)$

As vertex and focus are $5 - 3 = 2$ units apart, then $p = 2$ equation is

${\left(y + 1\right)}^{2} = - 8 \left(x - 5\right)$ or $x = - \frac{1}{8} {\left(y + 1\right)}^{2} + 5$

graph{x=-1/8(y+1)^2+5 [-21, 19, -11, 9]}