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?

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Answer 1 · 2017-05-30T08:17:50

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

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.