Find the conic section represented by the equation: x=y^2+10
I think this is a parabola. But how do you tell if the equation is a hyperbola or not?
Put it in the standard form $\displaystyle Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.
Compute $\displaystyle B^2 - 4AC$: if the number is negative, it's an ellipse; if it's zero, it's a parabola; if it's positive, it's a hyperbola.
Here $\displaystyle B^2 - 4AC = 0^2 - 4(0)(1) = 0$
Without the "xy" term it is particularly easy to tell:
If there is either $\displaystyle x^2$ or $\displaystyle y^2$ but not both, the graph is a parabola.
If the coefficients of $\displaystyle x^2$ and $\displaystyle y^2$ (when they are on the same side of the equation) are of opposites sign, it is hyperbola.
If the coefficients of $\displaystyle x^2$ and $\displaystyle y^2$ (when they are on the same side of the equation) are of the same sign, it is an ellipse.
Special case: if the coefficients are exactly the same, then the graph is a circle.
You can tell iff an equation is a hyperbola, ellipse, or parabola by making the right side of the equation 1, analyzing the coefficients of the variable terms, and then observing whether or not the sign seperating the terms is positive or negative.
In this case...move y^2 to the other side
$\displaystyle x-y^2=10$
Divide both sides by ten to analyze the coefficients
$\displaystyle \frac{1}{10}x-\frac{1}{10}y^2=1$
Since the coefficients are equal and only one of the variable terms is quadratic, this is a parabola.
A = 1
B = -4
C = -20
$\displaystyle B^2 - 4AC = (-4)^2 - 4(1)(-20) = 16 + 80 = 96$
Probably this is in an upcoming chapter for you, but the xy term can be "rotated out" of the equation using a substitution. Doesn't matter, though. The $\displaystyle B^2 - 4AC$ catches them every time. I guess you could say it helps you 'discriminate' between the types ... which is why it's called the 'discriminant.'