Find the vertex, focus, and directrix of the parabola given by the following equation:

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

I get up to

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

I don't know how to finish this.

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- Jun 29th 2010, 10:18 PMRBlaxParabola: Vertex, Focus, Directrix
Find the vertex, focus, and directrix of the parabola given by the following equation:

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

I get up to

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

I don't know how to finish this. - Jun 30th 2010, 04:48 AMHallsofIvy
**Don't**do that!

Instead of factoring, complete the square.

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

$\displaystyle 3x+ y^2+ 8y+ 4= 0$ becomes $\displaystyle 3x= -y^2- 8y- 4= -(y^2+ 8y+ 4)$

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

$\displaystyle 3x= -(y^2+ 8x+4)= -(y^2+ 8x+ 16- 16+ 4)= -((y+4)^2- 12)$

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

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

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