How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola #y^2=36+4x^2#?

1 Answer
Jul 20, 2017

Answer:

See answer below.

Explanation:

The standard form of a hyperbola is either

#x^2/a^2-y^2/b^2=1#

or

#color(blue)(y^2/a^2-x^2/b^2=1)#

Our hyperbola is of the second form and has a vertical transverse axis, which means the foci and vertices are on the y-axis.

#color(red)("Short cuts"):#

Vertices will be at #(0,a)# and #(0,-a)#

Asymptotes will be #y=+-(a/b)x#

Foci are at #(0,+-c)#, where #c=sqrt(a^2+b^2)#

So you can plug #a# and #b# into the above formulae, but I will give a bit of an explanation below.

First, rewrite the equation in standard form so you can find #a# and #b#:

#y^2=36+4x^2rArry^2-4x^2=36rArry^2/6^2-x^2/3^2=1#

#:.a=6# and #b=3#

1) The vertices are the turning points. These are found by setting x and y equal to zero.

y-intercepts (set x = 0):

#y^2=36+4(0)^2rArry=+-6#

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:

#(0,6)# and #(0,-6)#

2) For the asymptotes, solve the equation for y and look at the behaviour as x approaches #+-oo#. For large positive and negative values of x the hyperbola is essentially behaving like a straight line. I.e. it is asymptoting towards straight lines.

#y^2=36+4x^2rArry=+-sqrt(36+4x^2)#

As x gets really large, we can ignore the 36, as it is essentially zero when compared with #oo#. The equation then becomes

#y=+-sqrt(4x^2)=+-2x#

These are now the equations of the asymptotes:

#y=2x#

#y=-2x#

3) Foci equation:

#a^2+b^2=c^2#

Solve for c to find the y-coordinates:

#c=+-sqrt(a^2+b^2)=+-sqrt(6^2+3^2)=+-sqrt(45)=+-3sqrt(5)#

Foci coordinates:

#(0,3sqrt5)# and #(0,-3sqrt5)#

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 #+-oo# it asymptotes towards the two straight lines.

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