# What is the equation of the parabola with a focus at (7,5) and a directrix of y= -3?

Mar 30, 2018

Parabola's equation is $y = \frac{1}{16} {\left(x - 7\right)}^{2} + 1$ and vertex is $\left(7 , 1\right)$.

#### Explanation:

Parabola is locus of a point which moves so that its distance from a given point calld focus and a given line ccalled directrix is always constant.

Let the point be $\left(x , y\right)$. Here focus is $\left(7 , 5\right)$ and distance from focus is $\sqrt{{\left(x - 7\right)}^{2} + {\left(y - 5\right)}^{2}}$. Its distance from directrix $y = - 3$ i.e. $y + 3 = 0$ is $| y + 3 |$.

Hence equaion of parabola is

(x-7)^2+(y-5)^2)=|y+3|^2

or ${x}^{2} - 14 x + 49 + {y}^{2} - 10 y + 25 = {y}^{2} + 6 y + 9$

or ${x}^{2} - 14 x + 65 = 16 y$

i.e. $y = \frac{1}{16} \left({x}^{2} - 14 x + 49 - 49\right) + \frac{65}{16}$

or $y = \frac{1}{16} {\left(x - 7\right)}^{2} + \frac{65 - 49}{16}$

or $y = \frac{1}{16} {\left(x - 7\right)}^{2} + 1$

Hence parabola's equation is $y = \frac{1}{16} {\left(x - 7\right)}^{2} + 1$ and vertex is $\left(7 , 1\right)$.

graph{(1/16(x-7)^2+1-y)((x-7)^2+(y-1)^2-0.15)((x-7)^2+(y-5)^2-0.15)(y+3)=0 [-12.08, 27.92, -7.36, 12.64]}