# Parabola: Vertex, Focus, Directrix

#### RBlax

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.

#### HallsofIvy

MHF Helper
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 $$\displaystyle (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\)