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

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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) About Me  (Show Source):
You can put this solution on YOUR website!
%28y-3%29%5E2%2F16+-+%28x%2B2%29%5E2%2F9=1
  • 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.
  • a%5E2 is always under the first term for a (vertical or horizontal) hyperbola. So a%5E2+=+16 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.
  • b%5E2 is always under the second term for a (vertical or horizontal) hyperbola. So b%5E2+=+9 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:
    %28y-3%29%5E2%2F16+-+%28%28-1%29%2B2%29%5E2%2F9=1
    %28y-3%29%5E2%2F16+-+%281%29%5E2%2F9=1
    %28y-3%29%5E2%2F16+-+1%2F9=1
    Multiplying by 144 (which is the lowest common denominator of 16 and 9):
    9%28y-3%29%5E2+-+16=144
    Adding 16:
    9%28y-3%29%5E2+=160
    Dividing by 9:
    %28y-3%29%5E2+=160%2F9
    Square root of each side:
    y-3+=+sqrt%28160%2F9%29 or y-3+=+-sqrt%28160%2F9%29
    Rationalizing the denominator:
    y-3+=+sqrt%28160%29%2F3 or y-3+=+-sqrt%28160%29%2F3
    Adding 3:
    y+=+3+%2B+sqrt%28160%29%2F3 or y+=+3+-+sqrt%28160%29%2F3
    Using our calculators to find a decimal approximation for these (rounded to two places) we get:
    y+=+7.27 or y+=+-1.27
    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 c%5E2=a%5E2-b%5E2. Using the values we found for a and b we get: c%5E2+=+16+-+9+=+7. So c+=+sqrt%287%29. 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 + sqrt%287%29) and (-2, 3 - sqrt%287%29)
    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.)