SOLUTION: What is the graph of a hyperbola having the equation of (y-3)^2/16

SOLUTION: What is the graph of a hyperbola having the equation of (y-3)^2/16 - (x+2)^2/9=1

Question 717552: What is the graph of a hyperbola having the equation of (y-3)^2/16 - (x+2)^2/9=1
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

The first thing we can determine from the equation is the center: (-2, 3)
Second, with the y-term in front, this will be a vertically-oriented hyperbola. In other words, the transverse axis is vertical.
is always under the first term for a (vertical or horizontal) hyperbola. So which means that a is 4.
With vertical transverse axis we will go up and down, by a distance of a, from the center to reach the vertices. To go vertically we add or subtract from the y-coordinate. So the vertices are:
(-2, 3+4) or (-2, 7)
and
(-2, 3-4) or (-2, -1)
Plot these two points.
is always under the second term for a (vertical or horizontal) hyperbola. So making b = 3.
In a vertical hyperbola, the a runs vertically and the b runs horizontally. Since slope is rise/run, the slope of the asymptotes for this hyperbola will be +a/b. Using the values we have for a and b we get +4/3 for the slopes of the asymptotes.
The asymptotes pass through the center. So we can sketch in the asymptote with a slope of 4/3 by going up 4 and then to the right 3. This puts us at the point (1, 7). Draw a dotted line that passes through this point and the center. For the asymptote with a slope of -4/3, go up 4 and to the left 3. This puts us at the point (-5, 7). Draw another dotted line that passes through this point and the center.
With the vertices and the asymptotes we could already draw a rough sketch of the hyperbola. If you want a more accurate graph, start building a table of values. As you do so, take advantage of the symmetry of hyperbolas. (Hyperbolas are symmetric to both the transverse axis and the conjugate axis.) For example, let's pick a value of -1 for x:

Multiplying by 144 (which is the lowest common denominator of 16 and 9):

Adding 16:

Dividing by 9:

Square root of each side:
or
Rationalizing the denominator:
or
Adding 3:
or
Using our calculators to find a decimal approximation for these (rounded to two places) we get:
or
The two y values reflect the symmetry about the conjugate axis. To take advantage of the symmetry about the transverse axis, we note that the x value we used, -1, is exactly one unit to the right of the transverse axis. Because of the symmetry, the an x value that is one to the left of the transverse axis, -2-1 or -3, will have the same y values as the points that are 1 to the right. So by using the symmetry and one x value, -1, we can get 4 different points:
(-1, 7.27) (-1, -1.27) (-3, 7.27) (-3, -1.27)
Plot these points. With these 4 points and the two vertices, maybe we have enough points. If not, pick another x value, find four more points and repeat until you are satisfied that you can see the path of the hyperbola.
Although not part of the graph, often one is asked to find the foci, too. The distance from the center to a focus is "c". And . Using the values we found for a and b we get: . So . The foci are also on the transverse axis so we will go up and down from the center to get to the foci:
(-2, 3 + ) and (-2, 3 - )
These are the exact coordinates for the two foci. I'll leave it up to you and your calculator if you want/need decimal approximations for the y coordinates.

FWIW, here is what the graph looks like. (The dotted lines of the asymptotes are not drawn because I do not know how to make algebra.com to draw a dotted line. And the two parts of the hyperbola are different colors because I had to tell algebra.com to draw them as separate graphs.)

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