# Parabola: Vertex, Focus, Directrix

• Jun 29th 2010, 10:18 PM
RBlax
Parabola: Vertex, Focus, Directrix
Find the vertex, focus, and directrix of the parabola given by the following equation:

$3x+y^2+8y+4=0$

I get up to

$y(y+8)=-3(x-4)$

I don't know how to finish this.
• Jun 30th 2010, 04:48 AM
HallsofIvy
Don't do that!

Instead of factoring, complete the square.

First, since the standard form for a parabola is either $y= (x-a)^2+ b$ or $x= (y- a)^2+ b$, your objective is to get it into that form- and that " $(y- a)^2$ should tell you that you want to get a "perfect square". First swap that y over to the other side of the equation:
$3x+ y^2+ 8y+ 4= 0$ becomes $3x= -y^2- 8y- 4= -(y^2+ 8y+ 4)$

Now, you should have learned that [tex](y+ a)^2= y^2+ 2ay+ a^2[/itex]. Compare that to " $y^2+ 8y$". Clearly that has 2a= 8 so a= 4. Then $a^2= 16$: $(y+ 4)= y^2+ 2(4)y+ (4)^2= y^2+ 8y+ 16$. To make $y^2+ 8y$ a "perfect square" (complete the square) we must add (and subtract) 16.

$3x= -(y^2+ 8x+4)= -(y^2+ 8x+ 16- 16+ 4)= -((y+4)^2- 12)$
so that, finally, $x= -\frac{1}{3}(y+4)^2+ \frac{1}{3}(12)= -\frac{1}{3}(y+ 4)^2+ 4$.

Okay, what does that "standard form" tell you about "vertex", "focus", and "directrix"?

Now, look at the formulas in your book. What does