Find the vertices, foci, and the equations of the asymptotes for the hyperbola..

INCOMING EQUATION

Question

Find the vertices, foci, and the equations of the asymptotes for the hyperbola..

INCOMING EQUATION

anonymous · Answer

$$x^{2}\div36 - y^{2}\div25 = 1$$

anonymous · Answer

So the foci. That's a tricky one. Why are there more than one?

dumbcow · Answer

like the name....and the song

ok since the x term is positive the hyperbola opens left/right
a = 6
b = 5

center: (0,0)
vertices : (+-6 , 0)
to find foci, solve for c
a^2 + b^2 = c^2
36 +25 = c^2
c = sqrt61

foci: (+-sqrt61 , 0)

asymptotes:
y = +- (b/a)x
y = +-(5/6)x

dumbcow · Answer

there are 2 foci because a hyperbola has 2 parabolas opening in opposite directions
thus 2 vertices and 2 foci

anonymous · Answer

The vertice is 6 because square root of 36 is 6 right? That's how you determine that?

dumbcow · Answer

yes , the numbers underneath represent a^2 and b^2

anonymous · Answer

so why isn't it (6,5)

dumbcow · Answer

great question, a is always associated with the x_term here is a reference http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx

dumbcow · Answer

just remember : 
a goes with x  determines horizontal vertices/foci
b goes with y determines vertical vertices/foci