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

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.

meme