Question 714369: Hello, i need some guidance on this hyperbola problem. On my test, it says to find the vertices, center and the foci of y^2 over 5, minus x^2 over 11=1. Or, if this is another way to write out my problem-> y^2/5-x^2/11=1. Thank you
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! find the vertices, center and the foci of y^2 over 5, minus x^2 over 11=1.
y^2/5-x^2/11=1
This is an equation of a hyperbola with vertical transverse axis:
Its standard form: , (h,k)=)x,y) coordinates of center.
..
For given equation:
center:(0,0)
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a^2=5
a=√5
vertices: (0,0±a)=(0,0±√5)=(0,-√5), (0,+√5)
..
b^2=11
..
c^2=a^2+b^2=5+11=16
c=√16=4
Foci: (0,0±c)=(0,0±4)=(0,-4), (0,+4)
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