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Question 721572: What are the center, vertices, foci, and asymptotes of the hyperbola with equation (y+1)^2/36 - (x-2)^2/64 = 1?
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! The standard forms for the equation of a hyperbola are:
 for horizontal hyperbolas
and
 for vertical hyperbolas
In these standard forms...- The "h" and "k" are the x and y coordinates of the center of the hyperbola.
- The "a" is the distance from the center to a vertex. "a" is also used in finding "c" and in finding the slopes of the asymptotes of the hyperbola.
- The "b" is also a number used in finding "c" and in finding the slopes of the asymptotes of the hyperbola.
With the "y" before the subtraction and the "x" after, your hyperbola is a vertical one. So from this point on, we will be using:
Our first task will be to transform your equation into the standard form. Your equation
is already pretty close. But we want subtractions within the parentheses in the numerators. So we need to rewrite the first numerator as a subtraction:
And we need perfect squares in the denominators:
Now that the equation is in standard form we can read the h, k, a and b and find the desired components of the hyperbola:- h = 2 and k = -1 so the center is (2, -1).
- a = 6 so the distance from the center to a vertex is 6 units. Since the hyperbola is a vertical one, we want to go 6 units up and down from the center (instead of to the left and right). To go up and down 6 units from (2, -1) we will add and subtract 6 from the y-coordinate of the center:
- (2, -1+6) = (2, 5)
- (2, -1-6) = (2, -7)
So the two vertices are (2, 5) and (2, -7)- b = 8
- Now that we have a and b, we can:
- Find "c". For this we use the equation which connects the values of a, b and c:
Substituting in our values for a and b (actually we can just use  and  from the standard form):
 (We disregard the negative square root because "c" is also a distance)
c is the distance from the center to each focus. Again we go up and down from the center (like the vertices):
(2, -1+10) = (2, 9)
(2, -1-10) = (2, -11)
So the foci are: (2, 9) and (2, -11) - Find the slopes of the asymptotes. The slopes will be +(rise/run). Since the hyperbola is vertical, we went up and down a distance of "a" to get to the vertices, the "a" represents the the rise. This makes the "b" the run. So the slopes of the asymptotes are:
+(6/8) = +(3/4)
- The only thing left is to find the equations of the asymptotes. For this we use the slopes we have found and the fact that both asymptotes pass through the center. There are different ways to find the equation of a line when you have its slope and one of the points on the line. The simplest is to use the point-slope form:
where "m" is the slope and the  and  are the coordinates of a point on the line. If you are not familiar with this form, then just use what you have learned for finding the equation of a line when you have its slope and one of the points on the line. Using the point-slope form we get one asymptote with:
or
and the second asymptote with:
or
The problem, as posted, does not specify a certain form for the equations of the asymptotes. So these equations may be acceptable. But sometimes problems specify or teachers expect equations of lines in a certain form. Most common are:
Slope-intercept form: y = mx + b (Note: This "b" is not the same as the "b" for the hyperbola!)
or
Standard form: Ax + By = C (Note: The A, B and C are not the same as the a, b and c from the hyperbola!)
If you need the equations in one of these forms, just use algebra to transform
and
into the desired form.
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