Suppose the point P has Cartesian coordinates $\left({x}_{P}comma;{y}_{P}\right)$ and l is a horizontal line with the equation $yequals;{y}_{l}$ . We can do this without loss of generality by simply rotating the coordinate system appropriately. Now choose any point in the plane with the general coordinates $\left(x\,y\right)$.

The distance from P to $\left(x\,y\right)$ is given by ${d}_{P}\=\sqrt{{\left(x-{x}_{P}\right)}^{2}\+{\left(y-{y}_{P}\right)}^{2}}$.

The distance from l to $\left(x\,y\right)$ is given by ${d}_{l}\=\sqrt{{\left(x-{x}_{}\right)}^{2}\+{\left(y-{y}_{l}\right)}^{2}}\=\sqrt{{\left(y-{y}_{l}\right)}^{2}}$.

So, equating these distances and solving for y, we find:

$\sqrt{{\left(y-{y}_{l}\right)}^{2}}equals;\sqrt{{\left(x-{x}_{P}\right)}^{2}plus;{\left(y-{y}_{P}\right)}^{2}}$

${\left(y-{y}_{l}\right)}^{2}equals;{\left(x-{x}_{P}\right)}^{2}plus;{\left(y-{y}_{P}\right)}^{2}$

${y}^{2}-\left(2\cdot {y}_{l}\right)\cdot yplus;{y}_{l}^{2}equals;{x}^{2}-\left(2\cdot {x}_{P}\right)\cdot xplus;{x}_{P}^{2}plus;{y}^{2}-\left(2\cdot {y}_{P}\right)\cdot yplus;{y}_{P}^{2}$${}$

$-\left(2\cdot {y}_{l}\right)\cdot yplus;\left(2\cdot {y}_{P}\right)\cdot yequals;{x}^{2}-\left(2\cdot {x}_{P}\right)\cdot xplus;{x}_{P}^{2}plus;{y}_{P}^{2}-{y}_{l}^{2}$

$yequals;\frac{{x}^{2}-\left(2\cdot {x}_{P}\right)\cdot xplus;\left({x}_{P}^{2}plus;{y}_{P}^{2}-{y}_{l}^{2}\right)}{\left(-2\cdot {y}_{l}plus;2\cdot {y}_{P}\right)}$

Since all of the indexed variables are constants, this is simply a quadratic equation, proving that the locus of points equidistant from P and l forms a parabola.${}$