Question

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

Answer 1

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.