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.
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