Parabola: Vertex, Focus, Directrix

Jun 2010
16
0
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
Apr 2005
20,249
7,909
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\)