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nickyd63 · Feb 1, 2012, 07:46 PM

2x^2 2x 2y^2-6y-11=0 how do I find out what the conic equation is?

corrigan · Feb 1, 2012, 10:52 PM

I'm not exactly sure what you mean by "what the conic equation is". I think you mean what type of curve it is, also called the conic section. The reason it's called a conic section is because it's obtained by taking a conical surface (looks kind of like an hour glass that stretches to infinity) and intersecting it with a plane. The conic section is the curve formed by the intersection. The types of conics and their respective general formulas are the following:

An ellipse: $\frac{(x-h)^2}{a^2}+ \frac{(y-k)^2}{b^2} =1$ .

A circle: $(x-h)^2+ (y-k)^2 =r^2$ . Notice this is just a special type of ellipse when $a^2=b^2=r^2$ .

A parabola: $q\cdot x^2 =y$ or $q\cdot y^2 =x$ .

A Hyperbola: $\frac{(x-h)^2}{a^2}- \frac{(y-k)^2}{b^2} =1$ or $\frac{(y-k)^2}{a^2}- \frac{(x-h)^2}{b^2} =1$ .

After a bit of experience you can just look at an equation and tell what type. But I doubt your teacher wants you to just say "I eyeballed it". So what you need to do is manipulate the equation until it looks like one of the general formulas above. The important trick you need to remember when manipulating it is completing the square, which is:

$x^2+Bx=x^2+Bx + (\frac{B}{2})^2 - (\frac{B}{2})^2=(x + \frac{B}{2})^2 - \frac{B^2}{4}$

So let's look at you specific problem and see what we can do:

$2x^2+2x+2y^2-6y-11=0$

$\Rightarrow x^2+x+y^2-3y-\frac{11}{2}=0$ (I just divided everything by 2).

$\Rightarrow x^2 + x + (\frac{1}{2})^2 - (\frac{1}{2})^2 + y^2 - 3y + (\frac{-3}{2})^2 - (\frac{-3}{2})^2 - \frac{11}{2}=0$

$\Rightarrow (x^2 + x + (\frac{1}{2})^2) + (y^2 - 3y + (\frac{-3}{2})^2) - \frac{1}{4} - \frac{9}{4} - \frac{22}{4} =0$

$\Rightarrow (x + \frac{1}{2})^2 + (y-\frac{3}{2})^2- \frac{32}{4} =0$

$\Rightarrow (x + \frac{1}{2})^2 + (y-\frac{3}{2})^2 = 8$

So it's a circle. If you're trying to draw this, the center will be at $(-\frac{1}{2}, \frac{3}{2})$ and the radius will be $\sqrt{8} = 2 \sqrt{2}$ .