# Thread: Focus, directrix & vertex of parabola

1. ## Focus, directrix & vertex of parabola

Hi

Find the focus, directrix and vertex of the parabola x = $y^2 - 6y + 3$.

Focus ( $-5\frac{3}{4}$, 3) OR (-5, 3)

Directrix x = - 6 $\frac{1}{4}$ OR x = - 7

Vertex (-6, 3)

If the answers are the former, could someone please show me how to do this question?

Thanx

$\sqrt{\frac{d^2}{n}}$

2. Originally Posted by xwrathbringerx
Hi

Find the focus, directrix and vertex of the parabola x = $y^2 - 6y + 3$.

Focus ( $-5\frac{3}{4}$, 3) OR (-5, 3)

Directrix x = - 6 $\frac{1}{4}$ OR x = - 7

Vertex (-6, 3)

If the answers are the former, could someone please show me how to do this question?

Thanx
Use the formulae here Parabola - Wikipedia, the free encyclopedia but interchange x and y ....

the given equation is x=y^2-6y+3
it can be written as
x=(y-3)^2 -6
or x+6=(y-3)^2
now shift the coordinate system such that x+6=X and y-3=Y
then the equation becomes Y^2=X
compare it with Y^2=4aX
we get a=1/4

Vertex X=o,Y=o but in new coordinate system i.e. XY system
or x+6=0 ,y-3=0 in orig. coordinate system i.e. xy system
i.e. (-6,3)

now

Focus=(1/4,0) but in new coordinate system i.e.XY system
or X=1/4
or x+6=1/4
or x=1/4-6= -23/4

similarly Y=0 gives y=3
so the focus is (, 3)

we know that the equation of diretrix is X=-a
so, diretrix is
X=-1/4 but in new coordinate system i.e. XY system

or x+6=-1/4 in orig. system i.e. xy system

or x=-25/4

.