Asked by Anonymous
                What are the vertices and co-vertices of y^2/25 - (x + 6)^2/144 = 1
            
            
        Answers
                    Answered by
            Reiny
            
    standard notation:
(x+6)^2/144 - y^2/25 = -1
so the hyperbola is vertical and its centre is (-6,0)
a^2 = 25 and b^2 = 144
a = ±5 and b = ± 12
vertices are 12 up and 12 down from (-6,0) which is
(-6,12) and (-6,-12)
If I recall the co-vertices, although not points on the hyperbola would be 5 to the right and 5 to the left of (-6,0), which would be
(-1,0) and (-11,0)
    
(x+6)^2/144 - y^2/25 = -1
so the hyperbola is vertical and its centre is (-6,0)
a^2 = 25 and b^2 = 144
a = ±5 and b = ± 12
vertices are 12 up and 12 down from (-6,0) which is
(-6,12) and (-6,-12)
If I recall the co-vertices, although not points on the hyperbola would be 5 to the right and 5 to the left of (-6,0), which would be
(-1,0) and (-11,0)
                    Answered by
            Anonymous
            
    But the graph of the original equation shows the vertices to be (-6, 5) and (-6, -5).
    
                    Answered by
            Reiny
            
    OF COURSE!!
got my a's and b's reversed.
I should have said:
a^2 = 144 and b^2 = 25
giving a = ±12 and b = ±5
so I would go up 5 and down 5 from the centre
so (-6,5) and (-6,-5)
the same would be true for the co-vertices, should be
(6,0) and (-18,0)
http://www.wolframalpha.com/input/?i=plot+%28x%2B6%29%5E2%2F144+-+y%5E2%2F25+%3D+-1+
    
got my a's and b's reversed.
I should have said:
a^2 = 144 and b^2 = 25
giving a = ±12 and b = ±5
so I would go up 5 and down 5 from the centre
so (-6,5) and (-6,-5)
the same would be true for the co-vertices, should be
(6,0) and (-18,0)
http://www.wolframalpha.com/input/?i=plot+%28x%2B6%29%5E2%2F144+-+y%5E2%2F25+%3D+-1+
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