# Thread: General Equation of a Conic

1. ## General Equation of a Conic

The general equation of a conic $\displaystyle Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ can tell you the kind of conic it defines even when $\displaystyle B \neq 0$. The following rule applies:
Let $\displaystyle K= B^2-4AC$, then the equation defines
an ellipse if $\displaystyle K < 0$.
a parabola if $\displaystyle K = 0$.
a hyperbola if $\displaystyle K > 0$.

Tell what kind of conic each equation represents. On your own graph paper, graph each function by solving for $\displaystyle y$. (Hint: To do this, you may need the quadratic formula, regarding $\displaystyle y$ as the variable and $\displaystyle x$ as part of the coeffiecient of $\displaystyle y$.

(a) $\displaystyle 2x^2 - xy - 10 =0$
Conic: hyperbola
y= ________________

(b) $\displaystyle x^2 + xy +y^2 - 8 = 0$
Conic: ellipse
y= ________________

(c) $\displaystyle x^2 + 2xy + y^2 - 2x = 0$
Conic: parabola
y= ________________
I used $\displaystyle K= B^2-4AC$ to find what type of conic it is for each equation, (a) being a hyperbola, (b) being an ellipse, and (c) being a parabola. The problem I'm having is how do I put the equation in $\displaystyle y =$ from so I can graph it on graph paper. The hint told me to use the quadratic formula, but it's vague on how to input the values of the equation. So can someone point me in the right way to how to put the equations in $\displaystyle y=$ using the quadratic formula so I can graph it on graph paper?

2. Originally Posted by chrozer
I used $\displaystyle K= B^2-4AC$ to find what type of conic it is for each equation, (a) being a hyperbola, (b) being an ellipse, and (c) being a parabola. The problem I'm having is how do I put the equation in $\displaystyle y =$ from so I can graph it on graph paper. The hint told me to use the quadratic formula, but it's vague on how to input the values of the equation. So can someone point me in the right way to how to put the equations in $\displaystyle y=$ using the quadratic formula so I can graph it on graph paper?
For the 1st equation you can isolate y just solve for it.

In the 2nd I like to complete the square so here we go....

$\displaystyle y^2+xy=8-x^2$ so take half the middle coeffeint x and square it and add it to both sides.

$\displaystyle y^2+xy+\frac{x^2}{4}=8-x^2+\frac{x^2}{4}$ completing the square we get..

$\displaystyle (y+\frac{x}{2})^2=8-\frac{3x^2}{4}$

solving for y we get..

$\displaystyle y+\frac{x}{2}=\pm \sqrt{8-\frac{3x^2}{4}}$

$\displaystyle y= - \frac{x}{2}\pm \sqrt{8-\frac{3x^2}{4}}$

3. Originally Posted by TheEmptySet
For the 1st equation you can isolate y just solve for it.

In the 2nd I like to complete the square so here we go....

$\displaystyle y^2+xy=8-x^2$ so take half the middle coeffeint x and square it and add it to both sides.

$\displaystyle y^2+xy+\frac{x^2}{4}=8-x^2+\frac{x^2}{4}$ completing the square we get..

$\displaystyle (y+\frac{x}{2})^2=8-\frac{3x^2}{4}$

solving for y we get..

$\displaystyle y+\frac{x}{2}=\pm \sqrt{3-\frac{3x^2}{4}}$

$\displaystyle y= - \frac{x}{2}\pm \sqrt{3-\frac{3x^2}{4}}$
Ok...thnx. But what did the hint meant by using the quadratic equation to solve for $\displaystyle y=$? I mean I know how to complete the square..but it gave a hint to use quadratic formula, and I assume you have to use that to find the $\displaystyle y=$ form of the equation.

4. Originally Posted by chrozer
Ok...thnx. But what did the hint meant by using the quadratic equation to solve for $\displaystyle y=$? I mean I know how to complete the square..but it gave a hint to use quadratic formula, and I assume you have to use that to find the $\displaystyle y=$ form of the equation.
You will actually get the same thing for example...

$\displaystyle y^2+xy+(x^2-8)=0$

so $\displaystyle a=1 \mbox{ } b= x \mbox{ } c=x^2-8$

so plugging into the quadratic fomula gives...

$\displaystyle y=\frac{-x \pm \sqrt{(x)^2-4(1)(x^2-8)}}{2}=\frac{-x \pm \sqrt{32-3x^2}}{2}$

$\displaystyle \frac{-x \pm \sqrt{32-3x^2}}{2}$
Is the same as the other answer by completing the square.

$\displaystyle y= - \frac{x}{2}\pm \sqrt{8-\frac{3x^2}{4}}=-\frac{x}{2} \pm \sqrt{\frac{32-3x^2}{4}}=\frac{-x \pm \sqrt{32-3x^2}}{2}$

Sorry about the above typo and any confustion it may have caused.

5. Originally Posted by TheEmptySet
You will actually get the same thing for example...

$\displaystyle y^2+xy+(x^2-8)=0$

so $\displaystyle a=1 \mbox{ } b= x \mbox{ } c=x^2-8$

so plugging into the quadratic fomula gives...

$\displaystyle y=\frac{-x \pm \sqrt{(x)^2-4(1)(x^2-8)}}{2}=\frac{-x \pm \sqrt{32-3x^2}}{2}$

$\displaystyle \frac{-x \pm \sqrt{32-3x^2}}{2}$
Is the same as the other answer by completing the square.

$\displaystyle y= - \frac{x}{2}\pm \sqrt{8-\frac{3x^2}{4}}=-\frac{x}{2} \pm \sqrt{\frac{32-3x^2}{4}}=\frac{-x \pm \sqrt{32-3x^2}}{2}$

Sorry about the above typo and any confustion it may have caused.
And I guess the only way of graphing it on graph paper is to use a calculator?

One more thing - Why does the $\displaystyle c$ value included both $\displaystyle x^2 - 8$? The problem says that $\displaystyle C$ is the coefficient of $\displaystyle y^2$ while $\displaystyle F = -8$.

Edit - NVM...i got it now. Thnx again.