# What is the vertex form of the equation of the parabola with a focus at (1,-9) and a directrix of y=-1 ?

Apr 9, 2017

$y = - \frac{1}{16} {\left(x - 1\right)}^{2} + 5$

#### Explanation:

Parabola is the locus of a point which moves so that its distance from a point called focus and a line called directrix is always same.

Hence a point, say $\left(x , y\right)$ on the desired parabola will be equidistant from focus $\left(1 , - 9\right)$ and directrix $y = - 1$ or $y + 1 = 0$.

As the distance from $\left(1 , - 9\right)$ is $\sqrt{{\left(x - 1\right)}^{2} + {\left(y + 9\right)}^{2}}$ and from $y + 1$ is $| y + 1 |$, we have

${\left(x - 1\right)}^{2} + {\left(y + 9\right)}^{2} = {\left(y + 1\right)}^{2}$

or ${x}^{2} - 2 x + 1 + {y}^{2} + 18 y + 81 = {y}^{2} + 2 y + 1$

or ${x}^{2} - 2 x + 16 y + 81 = 0$

or $16 y = - 1 \left({x}^{2} - 2 x + 1 - 1\right) - 81$

or $16 y = - - \left({x}^{2} - 2 x + 1\right) + 1 - 81$

or $y = - \frac{1}{16} {\left(x - 1\right)}^{2} + 5$

Hence, vertex is $\left(1 , - 5\right)$ and axis of symmetry is $x = 1$

graph{(y+1/16(x-1)^2+5)(y+1)(x-1)((x-1)^2+(y+9)^2-0.04)=0 [-20.08, 19.92, -17.04, 2.96]}