Conic Equation in "y =" form

[chrozer]

How do you put these conic equations in "y =" form?

2x^2 - xy - 10 = 0

x^2 + xy + y^2 = 0

x^2 + 2xy + y^2 + 2x = 0

It said on the worksheet to use the quadratic formula to put the equation in "y =" form by regarding "y" as the variable and "x" as part of the coeffiecient of "y".

 

[Loren]

Suppose we had the equation 4x^2+2y^2+xy+x=0

We arrange it in powers of x, the 2nd power of x coming first, the first power of x coming second and finally the zero power of x coming last.

4x^2 + xy + x +2y^2 = 0

We see we have two terms that are in the first power of x, so we rewrite the equation as...

4x^2 + (y+1)x + 2y^2 = 0

We remember the quadratic formula applies to a quadratic equation in the form ax^2 + bx + c = 0.

Applying that to our equation we see that a=4, b=(y+1) and c=2y^2.

So, we plug those values into the quadratic formula and simplify.
\(\displaystyle x=\frac{-(y+1)\pm\sqrt{(y+1)^2-4\cdot 4\cdot 2y^2}}{2\cdot 4}\)
\(\displaystyle x=\frac{-y-1\pm\sqrt{y^2+2y+1-32y^2}}{8}\)
\(\displaystyle x=\frac{-y-1\pm\sqrt{-31y^2+2y+1}}{8}\)

I think this is what you are after.