How do you graph the hyperbola #(x+1)^2/9-(y-3)^2/4=1# and find the center, lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes?

1 Answer
Jun 22, 2018

Answer:

See all the stuff below for an explanation!

Explanation:

graph{(((x+1)^2)/9)-(((y-3)^2)/4)=1 [-11.165, 8.835, -2.2, 7.8]}

This is what the graph of your equation looks like. Pretty cool, right? Let's find out how to get this.

First, let's get our center. For this, let's look at the equation we have. The center of a hyperbola will have the opposites of the two numbers after the #x# and #y# values, respectively. In our case, it will be #(-1,3)#.

We can also find our lines that have the transverse and conjugate axes. Since this is a horizontal hyperbola (#x# is found on the first fraction and #y# on the second), our transverse axis will be #y=3#. Our conjugate axis will be #x=-1#. Note that for a horizontal hyperbola, the transverse axis is horizontal (y=[number]) and the conjugate axis is vertical (x=[number]).

Next, let's find our #a, b,# and #c# values as these will help us find the rest of what we need. In a hyperbola equation, the #a# value will always be the denominator the first fraction and the #b# value will always be the denominator of the second. The values will always be squared, so you need to find the square roots for them:

#sqrt(9)=3, sqrt(4)=2#

#a= 3# and #b=2#. But what about c? To do that, let's use an equation you've probably used before:

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

Let's plug in our values.

#9+4=c^2#
#13=c^2#
#sqrt(13)=c#

So, we have our values. What next? Let's get our vertices, which are always #a# units away from the center in opposite directions. The vertices, in our case, will be 3 units to the left and right of the center #(-1+-3, 3)#. They will be #(-4,3)# and #(2,3)#.

The foci are also along the same line, but they are #c# units away #(-1+-sqrt(13), 3)#. It's fine if you write the foci this way.

Finally, let's get the asymptotes. For a horizontal hyperbola, the asymptotes are #y-k=+-b/a(x-h)# (Center is #(h, k)#). This means that our asymptotes are:

#y-3=+-2/3(x+1)#

So sorry for the long answer, but there was a lot of info to cover. Hope this helps!