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Sketch the graph for
Solution:
Consider the given equation above
The above equation represents a hyperbola because one of the 2nd degree variable is negative. If y2 is negative, then the curve of the hyperbola opens to the left and to the right. If x2 is negative, then the curve of the hyperbola opens upward and downward.
The first step is to reduce the given equation of the hyperbola to its standard form as follows
Group the above terms according to its variables
Complete the square of the grouped terms
Divide both sides of the equation by 36
Since y is negative, then the curve of the hyperbola opens to the left and to the right. The transverse axis is parallel to the x-axis where the center, two foci, and two vertices are located.
The center of the hyperbola is C(-1, 2).
The semi-major axis of the hyperbola is
Therefore, the main vertices of the hyperbola are
V1(-1 + 2, 2) = V1(1, 2)
V2(-1 - 2, 2) = V2(-3, 2)
The semi-minor axis of the hyperbola is
Therefore, the other vertices of the hyperbola are
V3(-1, 2 + 1½) = V3(-1, 3½)
V4(-1, 2 - 1½) = V4(-1, ½)
These two vertices of the hyperbola will not be used to draw the curve proper but it is helpful to draw the imaginary rectangle. To draw the imaginary rectangle, draw a light vertical line that passes through V1, another light vertical line that passes through V2, a light horizontal line that passes through V3, and another light horizontal line that passes through V4. The resulting imaginary rectangle will be a guide to draw the two asymptotes of the hyperbola in which you will connect and extend the two opposite vertices of a rectangle. The two asymptotes of the hyperbola should meet or intersect at the center of the hyperbola. The two equations of the asymptotes of the hyperbola are
Change the right side of the equation to zero, we have
Factor the above equation as the sum and the difference of two squares
Equate each factor to zero to get the equations of the two asymptotes.
For the first asymptote of the hyperbola
Multiply both sides of the equation by 6, we have
For the second asymptote of the hyperbola
Multiply both sides of the equation by 6, we have
Next, we need to get the coordinates of the foci and the ends of the latera recta of the hyperbola. First, we need to solve for the value of c which is the distance of the focus to the center of the hyperbola as follows
Therefore, the two foci of the hyperbola are
F1(-1 + 2½, 2) = F1(1½, 2)
F2(-1 - 2½, 2) = F2(-3½, 2)
Next, solve for the length of the latus rectum as follows
Therefore, the end points of the latera recta are
L1(1½, 2 + 1⅛) = L1(1½, 3⅛)
L2(1½, 2 - 1⅛) = L2(1½, ⅞)
L3(-3½, 2 + 1⅛) = L3(-3½, 3⅛)
L4(-3½, 2 - 1⅛) = L4(-3½, ⅞)
Finally, connect the points L3, V2, and L4 to draw a curve that opens to the left and connect the points L1, V1, and L2 to draw a curve that opens to the right as follows
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| Photo by Math Principles in Everyday Life |