# What is the equation in standard form of the parabola with a focus at (14,5) and a directrix of y= -15?

Dec 30, 2017

The equation of parabola is $y = \frac{1}{40} {\left(x - 14\right)}^{2} - 5$

#### Explanation:

Focus is at $\left(14 , 5\right)$and directrix is $y = - 15$. Vertex is at midway

between focus and directrix. Therefore vertex is at

$\left(14 , \frac{5 - 15}{2}\right) \mathmr{and} \left(14 , - 5\right)$ . The vertex form of equation of

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

$h = 14 \mathmr{and} k = - 5$ So the equation of parabola is

$y = a {\left(x - 14\right)}^{2} - 5$. Distance of vertex from directrix is

$d = 15 - 5 = 10$, we know $d = \frac{1}{4 | a |} \therefore | a | = \frac{1}{4 d}$ or

$| a | = \frac{1}{4 \cdot 10} = \frac{1}{40}$ . Here the directrix is below

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

$\therefore a = \frac{1}{40}$ Hence the equation of parabola is

$y = \frac{1}{40} {\left(x - 14\right)}^{2} - 5$
graph{1/40(x-14)^2-5 [-90, 90, -45, 45]} [Ans]

Dec 30, 2017

${\left(x - 14\right)}^{2} = 40 \left(y + 5\right)$

#### Explanation:

$\text{the standard form of a parabola in "color(blue)"translated form}$ is.

•color(white)(x)(x-h)^2=4p(y-k)

$\text{where "(h,k)" are the coordinates of the vertex}$

$\text{and p is the distance from the vertex to the focus}$

$\text{since the directrix is below the focus then the curve}$
$\text{opens upwards}$

$\text{coordinates of vertex } = \left(14 , \frac{5 - 15}{2}\right) = \left(14 , - 5\right)$

$\text{and } p = 5 - \left(- 5\right) = 10$

$\Rightarrow \Rightarrow {\left(x - 14\right)}^{2} = 40 \left(y + 5\right) \leftarrow \textcolor{red}{\text{equation of parabola}}$