# What is the equation, in standard form, for a parabola with the vertex (1,2) and directrix y=-2?

Nov 29, 2016

The equation of the parabola is (x-1)^2=16(y-2

#### Explanation:

The vertex is $\left(a , b\right) = \left(1 , 2\right)$

The directrix is $y = - 2$

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

Therefore,

$- 2 = 2 - \frac{p}{2}$

$\frac{p}{2} = 4$

$p = 8$

The focus is $\left(a , b + \frac{p}{2}\right) = \left(1 , 2 + 4\right) = \left(1 , 6\right)$

$b + \frac{p}{2} = 6$

$\frac{p}{2} = 6 - 2 = 4$

$p = 8$

The distance any point $\left(x , y\right)$ on the parabola is equidisdant from the directrix and the focus.

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

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

${y}^{2} + 4 y + 4 = {\left(x - 1\right)}^{2} + {y}^{2} - 12 y + 36$

$16 y - 32 = {\left(x - 1\right)}^{2}$

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

The equation of the parabola is

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

graph{(x-1)^2=16(y-2) [-10, 10, -5, 5]}