# What is the equation of a parabola with a focus at (-2, 6) and a vertex at (-2, 9)? What if the focus and vertex are switched?

Jul 21, 2018

The equation is $y = - \frac{1}{12} {\left(x + 2\right)}^{2} + 9$. The other equation is $y = \frac{1}{12} \left(x + 2\right) \cdot 2 + 6$

#### Explanation:

The focus is $F = \left(- 2 , 6\right)$ and the vertex is $V = \left(- 2 , 9\right)$

Therefore, the directrix is $y = 12$ as the vertex is the midpoint from the focus and the directrix

$\frac{y + 6}{2} = 9$

$\implies$, $y + 6 = 18$

$\implies$, $y = 12$

Any point $\left(x , y\right)$ on the parabola is equidistant from the focus and the directrix

$y - 12 = \sqrt{{\left(x + 2\right)}^{2} + {\left(y - 6\right)}^{2}}$

${\left(y - 12\right)}^{2} = {\left(x + 2\right)}^{2} + {\left(y - 6\right)}^{2}$

${y}^{2} - 24 y + 144 = {\left(x + 2\right)}^{2} + {y}^{2} - 12 y + 36$

$12 y = - {\left(x + 2\right)}^{2} + 108$

$y = - \frac{1}{12} {\left(x + 2\right)}^{2} + 9$

graph{(y+1/12(x+2)^2-9)(y-12)=0 [-32.47, 32.45, -16.23, 16.25]}

The second case is

The focus is $F = \left(- 2 , 9\right)$ and the vertex is $V = \left(- 2 , 6\right)$

Therefore, the directrix is $y = 3$ as the vertex is the midpoint from the focus and the directrix

$\frac{y + 9}{2} = 6$

$\implies$, $y + 9 = 12$

$\implies$, $y = 3$

$y - 3 = \sqrt{{\left(x + 2\right)}^{2} + {\left(y - 9\right)}^{2}}$

${\left(y - 3\right)}^{2} = {\left(x + 2\right)}^{2} + {\left(y - 9\right)}^{2}$

${y}^{2} - 6 y + 9 = {\left(x + 2\right)}^{2} + {y}^{2} - 18 y + 81$

$12 y = {\left(x + 2\right)}^{2} + 72$

$y = \frac{1}{12} {\left(x + 2\right)}^{2} + 6$

graph{(y-1/12(x+2)^2-6)(y-3)=0 [-32.47, 32.45, -16.23, 16.25]}