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Question 610053: What are the vertices of the hyperbola given by the equation (y^2/81)-(x^2/49)=1?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! With the equation y^2/81-x^2/49=1 or 
(I believe in parentheses, but you did not need those),
you realize that your hyperbola is symmetrical to either side of the y-axis
(because what's valid for any value of x is valid for -x)
and it is symmetrical to either side of the x-axis
(because what's valid for any value of y is valid for -y).
As a consequence, it must be centered at the origin.
You also realize that y=0 is impossible, because it would mean that a negative number equals 1:
-x^2/49=1
So your hyperbola does not touch the x-axis (where y=0).
Then it must graph as a smile-shaped curve above the x-axis, and the symmetrical "frown" below the x-axis, with vertices at (0,b) and (0,-b).
If you make x=0, you get y^2/81=1, and from there you find
y^2=81, and the coordinates of the vertices as y=9 and y=-9.
So the vertices are (0,9) and (0,-9).
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