How do you identity if the equation 3x^2-2y^2+32y-134=0 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
May 30, 2017

This is a hyperbola! noticing the negative y^2 term and the positive x^2 term.

Explanation:

This is a hyperbola in the general form, which means you must complete the square before graphing.

3x^2 - 2y^2 + 32y - 134 = 0

:. 3x^2 - 2(y^2 - 16y) = 134

:. 3x^2 - 2(y^2 - 16y + 64 - 64) = 134

:. 3x^2 - 2(y - 8)^2 + 128 = 134

:. 3x^2 - 2(y - 8)^2 = 6

:. (x^2) / 2 - ((y - 8)^2) / 3 = 1

This is the standard form for a hyperbolic graph, because it is of the form:
((x - h)^2) / a^2 - ((y - k)^2) / b^2 = 1

Of course, in this case h=0, k=8, a=sqrt2 and b=sqrt3.

The formula for the two oblique asymptotes are given by:

y = +-b/a(x-h) +k

:. y = +-sqrt3/sqrt2 * x + 8

:. y = (sqrt6x)/2 + 8 or y = -(sqrt6x)/2 +8.

Now you can sketch the two asymptotes on the plane.

After this, we must find any x- and y-intercepts, simply by letting one or the other equal zero.

Letting x=0, we get - 2y^2 + 32y - 134 = 0, a quadratic which can be solved easily using the quadratic formula.

Letting y=0, we get 3x^2 - 134 = 0, another quadratic.

Now you are ready to sketch this hyperbola. Remember not to direct the curve away from the asymptote as it approaches it.