What are the vertices and co-vertices of y^2/25 - (x + 6)^2/144 = 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)
But the graph of the original equation shows the vertices to be (-6, 5) and (-6, -5).
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+
To determine the vertices and co-vertices of the hyperbola given by the equation y^2/25 - (x + 6)^2/144 = 1, we can use the standard form of a hyperbola equation:
((y − k)^2) / a^2 - ((x − h)^2) / b^2 = 1
Comparing this with the given equation, we can see that the center of the hyperbola is at (-6, 0). The values of a and b correspond to the major and minor axes of the hyperbola, respectively.
To find the vertices and co-vertices, we need to determine the values of a and b.
For a hyperbola in this form, a is equal to the square root of the denominator under the y^2 term, and b is equal to the square root of the denominator under the (x + h)^2 term.
So, in this case, a = sqrt(25) = 5 and b = sqrt(144) = 12.
The vertices can be found by adding and subtracting 'a' from the center. So, the vertices are (-6, 0 ± a), which becomes (-6, -5) and (-6, 5).
The co-vertices can be found by adding and subtracting 'b' from the center. So, the co-vertices are (-6 ± b, 0), which becomes (-18, 0) and (6, 0).
Therefore, the vertices of the hyperbola are (-6, -5) and (-6, 5), and the co-vertices are (-18, 0) and (6, 0).