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