1. ## Parabola question

Find the equation of the parabola given that the focus is (6,0) and directrix is x=0.

No idea as we have had no class examples but I know about cartesian transformations.

Any help would be gratefully appreciated!

dojo

2. Originally Posted by dojo
Find the equation of the parabola given that the focus is (6,0) and directrix is x=0.

No idea as we have had no class examples but I know about cartesian transformations.

Any help would be gratefully appreciated!

dojo

Hi Dojo,

Here's one way to find it.

Your parabola will be in the form: $\displaystyle x=a(y-k)^2+h$ since it has a vertical directrex.

$\displaystyle x=0$

The dirctrex equation is $\displaystyle x=h-\dfrac{1}{4a}$

So, we have [1] $\displaystyle h-\dfrac{1}{4a}=0$

The focus is at $\displaystyle \left(h+\dfrac{1}{4a}\:,\:k\right)$

$\displaystyle F(6, 0)$ and $\displaystyle \boxed{k=0}$

So, we have [2] $\displaystyle h+\dfrac{1}{4a}=6$

Solve the system of equations [1] and [2] for h and a.

[1] $\displaystyle h-\dfrac{1}{4a}=0$

[2] $\displaystyle h+\dfrac{1}{4a}=6$

$\displaystyle 2h=6$

$\displaystyle \boxed{h=3}$

Substituting into [1],

$\displaystyle 3-\dfrac{1}{4a}=0$

$\displaystyle 3=\dfrac{1}{4a}$

$\displaystyle \boxed{a=\dfrac{1}{12}}$

Now, you will have everything you need to create your equation.

$\displaystyle x=\dfrac{1}{12}\left(y-0\right)^2+3$

$\displaystyle x=\dfrac{1}{12}y^2+3$