# What is the equation in standard form of the parabola with a focus at (1,4) and a directrix of y= 2?

Dec 6, 2015

$y = \frac{1}{4} {x}^{2} - \frac{1}{2} x + \frac{13}{4}$

#### Explanation:

If $\left(x , y\right)$ is a point on a parabola then
$\textcolor{w h i t e}{\text{XXX}}$the perpendicular distance from the directrix to $\left(x , y\right)$
is equal to
$\textcolor{w h i t e}{\text{XXX}}$the distance from $\left(x , y\right)$ to the focus.

If the directrix is $y = 2$
then
$\textcolor{w h i t e}{\text{XXX}}$the perpendicular distance from the directrix to $\left(x , y\right)$ is $\left\mid y - 2 \right\mid$

If the focus is $\left(1 , 4\right)$
then
$\textcolor{w h i t e}{\text{XXX}}$the distance from $\left(x , y\right)$ to the focus is $\sqrt{{\left(x - 1\right)}^{2} + {\left(y - 4\right)}^{2}}$ Therefore
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{\left\mid y - 2 \right\mid} = \sqrt{\textcolor{b l u e}{{\left(x - 1\right)}^{2}} + \textcolor{red}{{\left(y - 4\right)}^{2}}}$

color(white)("XXX")color(green)(y-2)^2) = color(blue)((x-1)^2)+color(red)((y-4)^2)

$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{\cancel{{y}^{2}} - 4 y + 4} = \textcolor{b l u e}{{x}^{2} - 2 x + 1} + \textcolor{red}{\cancel{{y}^{2}} - 8 y + 16}$

$\textcolor{w h i t e}{\text{XXX}} 4 y + 4 = {x}^{2} - 2 x + 17$

$\textcolor{w h i t e}{\text{XXX}} 4 y = {x}^{2} - 2 x + 13$

$\textcolor{w h i t e}{\text{XXX")y = 1/4x^2 -1/2x + 13/4color(white)("XXX}}$(standard form)
graph{1/4x^2-1/2x+13/4 [-5.716, 6.77, 0.504, 6.744]}