# Thread: Equation of a conic

1. ## Equation of a conic

Trying to figure out the whole "conic sections" thing... Here's the problem I'm stuck on: "Write the equation of the conic in standard form, and classify it as a circle, a parabola, an ellipse, or a hyperbola: 3x^2-2y^2+24x+12y+24=0"

Also, if anyone just knows any basic "steps" to follow when trying to find the foci, directerix, vertices, equations, etc, generally speaking, that would be IMMENSELY helpful! I've got a test tonight and I'm pretty confused!

Thanks in advance for any help!!!!!

2. ## Conics

A good first step is to complete the square on each of the variables.

$3x^2+24x-2y^2+12y=-24$

factoring and completing the square we get.

$3(x^2+8x+16)-2(y^2-6y+9)=-24+3(16)+(-2)(9)$

factoring and combining like terms

$3(x+4)^2-2(y-3)^2=6$

divide everything by 6 to put into standart form.

$\frac{(x+4)^2}{(\sqrt{2})^2}-\frac{(y-3)^2}{(\sqrt{3})^2}=1$

This is the standard form of a hyperbola.

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

The foci can be found using the equation c=foci

$c^2=b^2+a^2$ so the foci are at $c=\pm \sqrt{5}$

they are located at the points $(\pm \sqrt{5},0)$

The asymptotes of the hyperbola can be found with the equation

$y=\pm \frac{b}{a}x=\pm \frac{\sqrt{3}}{\sqrt{2}}$

Good luck.

P.S. The formula for foci is different for each conic section.