# What is the standard form of the equation of the parabola with a directrix at x=103 and a focus at (108,41)?

Nov 17, 2017

$x = \frac{1}{10} {\left(x - 41\right)}^{2} + \frac{211}{2}$

#### Explanation:

A parabola is the locus of a point, which moves so that its distance from a given line called directrix and a given point called focus, is always equal.

Now, the distance between two pints $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$ and distance of a point $\left({x}_{1} , {y}_{1}\right)$ from a line $a x + b y + c = 0$ is $| \frac{a {x}_{1} + b {y}_{1} + c}{\sqrt{{a}^{2} + {b}^{2}}} |$

Coming to parabola with directrix $x = 103$ or $x - 103 = 0$ and focus $\left(108 , 41\right)$, let the point equidistant from both be $\left(x , y\right)$. The distance of $\left(x , y\right)$ from $x - 103 = 0$ is

$| \frac{x - 103}{\sqrt{{1}^{2} + {0}^{2}}} | = | \frac{x - 103}{1} | = | x - 103 |$

and its distance from $\left(108 , 41\right)$ is

$\sqrt{{\left(108 - x\right)}^{2} + {\left(41 - y\right)}^{2}}$

and as the two are equal, equation of parabola would be

${\left(108 - x\right)}^{2} + {\left(41 - y\right)}^{2} = {\left(x - 103\right)}^{2}$

or ${108}^{2} + {x}^{2} - 216 x + {41}^{2} + {y}^{2} - 82 y = {x}^{2} + {103}^{2} - 206 x$

or $11664 + {x}^{2} - 216 x + 1681 + {y}^{2} - 82 y = {x}^{2} + 10609 - 206 x$

or ${y}^{2} - 82 y - 10 x + 2736 = 0$

or $10 x = {y}^{2} - 82 y + 2736$

or $10 x = {\left(y - 41\right)}^{2} + 1055$

or in vertex form $x = \frac{1}{10} {\left(x - 41\right)}^{2} + \frac{211}{2}$

and vertex is $\left(105 \frac{1}{2} , 41\right)$

Its graph appears as shown below, along with focus and directrix.

graph{(y^2-82y-10x+2736)((108-x)^2+(41-y)^2-0.6)(x-103)=0 [51.6, 210.4, -13.3, 66.1]}