Given the focus and directrix of a parabola, how do we find the equation of the parabola?

If we consider only parabolas that open upwards or downwards, then the directrix will be a horizontal line of the form $y=c$.

Let $\left(a,b\right)$ be the focus and let $y=c$ be the directrix. Let $\left({x}_{0},{y}_{0}\right)$ be any point on the parabola.

Any point, $\left({x}_{0},{y}_{0}\right)$ on the parabola satisfies the definition of parabola, so there are two distances to calculate:

- Distance between the point on the parabola to the focus
- Distance between the point on the parabola to the directrix

To find the equation of the parabola, equate these two expressions and solve for ${y}_{0}$.

Find the equation of the parabola in the example above.

Distance between the point $\left({x}_{0},{y}_{0}\right)$ and $\left(a,b\right)$:

$\sqrt{{\left({x}_{0}-a\right)}^{2}+{\left({y}_{0}-b\right)}^{2}}$

Distance between point $\left({x}_{0},{y}_{0}\right)$ and the line $y=c$:

$\left|{y}_{0}-c\right|$

(Here, the distance between the point and horizontal line is difference of their $y$-coordinates.)

Equate the two expressions.

$\sqrt{{\left({x}_{0}-a\right)}^{2}+{\left({y}_{0}-b\right)}^{2}}=\left|{y}_{0}-c\right|$

Square both sides.

${\left({x}_{0}-a\right)}^{2}+{\left({y}_{0}-b\right)}^{2}={\left({y}_{0}-c\right)}^{2}$

Expand the expression in ${y}_{0}$ on both sides and simplify.

${\left({x}_{0}-a\right)}^{2}+{b}^{2}-{c}^{2}=2\left(b-c\right){y}_{0}$

This equation in $\left({x}_{0},{y}_{0}\right)$ is true for all other values on the parabola and hence we can rewrite with $\left(x,y\right)$.

Therefore, the equation of the parabola with focus $\left(a,b\right)$ and directrix $y=c$ is

${\left(x-a\right)}^{2}+{b}^{2}-{c}^{2}=2\left(b-c\right)y$

**Example**:

If the focus of a parabola is $\left(2,5\right)$ and the directrix is $y=3$, find the equation of the parabola.

Let $\left({x}_{0},{y}_{0}\right)$ be any point on the parabola. Find the distance between $\left({x}_{0},{y}_{0}\right)$ and the focus. Then find the distance between $\left({x}_{0},{y}_{0}\right)$ and directrix. Equate these two distance equations and the simplified equation in ${x}_{0}$ and ${y}_{0}$ is equation of the parabola.

The distance between $\left({x}_{0},{y}_{0}\right)$ and $\left(2,5\right)$ is $\sqrt{{\left({x}_{0}-2\right)}^{2}+{\left({y}_{0}-5\right)}^{2}}$

The distance between $\left({x}_{0},{y}_{0}\right)$ and the directrix, $y=3$ is

$\left|{y}_{0}-3\right|$.

Equate the two distance expressions and square on both sides.

$\sqrt{{\left({x}_{0}-2\right)}^{2}+{\left({y}_{0}-5\right)}^{2}}=\left|{y}_{0}-3\right|$

${\left({x}_{0}-2\right)}^{2}+{\left({y}_{0}-5\right)}^{2}={\left({y}_{0}-3\right)}^{2}$

Simplify and bring all terms to one side:

${x}_{0}{}^{2}-4{x}_{0}-4{y}_{0}+20=0$

Write the equation with ${y}_{0}$ on one side:

${y}_{0}=\frac{{x}_{0}{}^{2}}{4}-{x}_{0}+5$

This equation in $\left({x}_{0},{y}_{0}\right)$ is true for all other values on the parabola and hence we can rewrite with $\left(x,y\right)$.

So, the equation of the parabola with focus $\left(2,5\right)$ and directrix is $y=3$ is

$y=\frac{{x}^{2}}{4}-x+5$