How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola #y^2=36+4x^2#?
See answer below.
The standard form of a hyperbola is either
Our hyperbola is of the second form and has a vertical transverse axis, which means the foci and vertices are on the y-axis.
Vertices will be at
Asymptotes will be
Foci are at
So you can plug
First, rewrite the equation in standard form so you can find
1) The vertices are the turning points. These are found by setting x and y equal to zero.
y-intercepts (set x = 0):
There are no real x-intercepts as setting y to zero leads you to the square root of a negative number.
So we have vertices at:
2) For the asymptotes, solve the equation for y and look at the behaviour as x approaches
As x gets really large, we can ignore the 36, as it is essentially zero when compared with
These are now the equations of the asymptotes:
3) Foci equation:
Solve for c to find the y-coordinates:
Now have a look at the graph, you can see that the foci and vertices are on the y-axis. You can also see that as x approaches