# How do you find the equation of the parabola with the given focus F(0,3) and directrix y=-5?

Oct 2, 2017

$y = \frac{1}{16} \cdot {x}^{2} - 1$

#### Explanation:

Since the directrix (the line on which the "bowl" of the parabola "floats" near to) is given by $y = - 5$ (a horizontal line), we know that the parabola opens vertically - either up or down.

The focus of a parabola lies "inside" of the "bowl" of the parabola, on the opposite side of the parabola as the directrix. This indicates that the parabola must open upwards from the directrix "towards" the focus.

The general form of an upward opening parabola is:

${\left(x - {x}_{0}\right)}^{2} = 4 f \left(y - {y}_{0}\right)$

The vertex $\left({x}_{0} , {y}_{0}\right)$ is located exactly in between the directrix line and the focus point. The distance between the two is 8, and so half of that distance is 4, making the vertex 4 units below the focus, or $\left(0 , - 1\right)$. This distance of 4 is also the "focus distance", or the distance between the vertex and the focus point, represented by $f$ in the formula above.

Putting this together, we arrive at:

${\left(x - 0\right)}^{2} = 4 \cdot 4 \cdot \left(y - \left(- 1\right)\right)$

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

$y = \frac{1}{16} \cdot {x}^{2} - 1$

graph{1/16*x^2-1 [-10, 10, -5, 5]}