# What is the equation in standard form of the parabola with a focus at (-18,30) and a directrix of y= 22?

Nov 12, 2017

The equation of parabola in standard form is
${\left(x + 18\right)}^{2} = 16 \left(y - 26\right)$

#### Explanation:

Focus is at $\left(- 18 , 30\right)$and directrix is $y = 22$. Vertex is at midway

between focus and directrix. Therefore vertex is at

$\left(- 18 , \frac{30 + 22}{2}\right)$ i.e at $\left(- 18 , 26\right)$ . The vertex form of equation

of parabola is y=a(x-h)^2+k ; (h.k) ; being vertex. Here

$h = - 18 \mathmr{and} k = 26$. So the equation of parabola is

$y = a {\left(x + 18\right)}^{2} + 26$. Distance of vertex from directrix is

$d = 26 - 22 = 4$, we know $d = \frac{1}{4 | a |}$

$\therefore 4 = \frac{1}{4 | a |} \mathmr{and} | a | = \frac{1}{4 \cdot 4} = \frac{1}{16}$. Here the directrix is below

the vertex , so parabola opens upward and $a$ is positive.

$\therefore a = \frac{1}{16}$ . The equation of parabola is $y = \frac{1}{16} {\left(x + 18\right)}^{2} + 26$

or $\frac{1}{16} {\left(x + 18\right)}^{2} = y - 26 \mathmr{and} {\left(x + 18\right)}^{2} = 16 \left(y - 26\right)$ or

${\left(x + 18\right)}^{2} = 4 \cdot 4 \left(y - 26\right)$.The standard form is

${\left(x - h\right)}^{2} = 4 p \left(y - k\right)$, where the focus is $\left(h , k + p\right)$

and the directrix is $y = k - p$. Hence the equation

of parabola in standard form is ${\left(x + 18\right)}^{2} = 16 \left(y - 26\right)$

graph{1/16(x+18)^2+26 [-160, 160, -80, 80]}