Question

What are the vertices and co-vertices of y^2/25 - (x + 6)^2/144 = 1

Answer 1

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)

Answer 2

But the graph of the original equation shows the vertices to be (-6, 5) and (-6, -5).

Answer 3

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+