Question 401080: Find the vertices of the hyperbola defined by this equation: (y+8)^2/36 - (x -7)^2/1=1.It tells me to put it in this form (x1,y1),(x2,y2).I have no idea How to do this,please help.I'd appreciate it :)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the vertices of the hyperbola defined by this equation: (y+8)^2/36 - (x -7)^2/1=1.It tells me to put it in this form (x1,y1),(x2,y2).I have no idea How to do this,please help.I'd appreciate it
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(y+8)^2/36 - (x -7)^2/1=1
This is a hyperbola with center at (7,-8) and the transverse axis is vertical.
Standard form of the hyperbola: (x-h)^2/a^2-(y-k)^2/b^2=1 (transverse axis horizontal) or (y-k)^2/a^2-(x-h)^2/b^2=1 (transverse axis vertical,like this case)
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a^2=36
a=6
b^2=1
b=1
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The vertex lies somewhere on the vertical transverse axis,so its x-coordinate is already determined to be 7. On this axis the vertices are 6 or "a" units above and below the center y-coordinate,-8.Therefore, the vertices are at (7,-2) and (7,-14)
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ans:(x,y) coordinates of the vertices are (7,-2) and (7,-14)
see the graph below to see what the hyperbola looks like:
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