# 36y^2-x^2=1 Finding the vertices and asymptotes to graph the hyperbola

• Nov 27th 2012, 10:33 AM
Ashz
36y^2-x^2=1 Finding the vertices and asymptotes to graph the hyperbola
Hi, I've also attached screenshot of the problem. I need help in solving this problem. I don't understand how to get from the equation 36y^2-x^2=1 to the format of ((y^2)/(a^2))-((x^2)/(b^2))=1, to determine what the vertices are.

• Nov 27th 2012, 12:40 PM
HallsofIvy
Re: 36y^2-x^2=1 Finding the vertices and asymptotes to graph the hyperbola
You can write it in the form $\frac{y^2}{\left(\frac{1}{6}\right)^2}- \frac{x^2}{1^2}= 1$. That's a hyperbola with center at the origin so its vertices are when one or the other variable is equal to 0. What do you get if y= 0? What do you get if x= 0?

An easy way to find the asymptotes is to argue that, for x very large, "1" will be small compared to either of the other terms and can be ignored. That is, for x large the graph will be approximately the same as $36y^2- x^2= (6y- x)(6y+ x)= 0$. That will be true as long as either 6y- x= 0 or 6y+ x= 0 which give straight lines.
• Nov 27th 2012, 12:56 PM
Ashz
Re: 36y^2-x^2=1 Finding the vertices and asymptotes to graph the hyperbola
Okay, I think I'm tracking ~

y=+-(1/6)x
Vertices (0, 1/6) & (0, -1/6)
Foci (0, sqrt 37/36) & (0, -sqrt 37/36)
• Nov 27th 2012, 12:58 PM
HallsofIvy
Re: 36y^2-x^2=1 Finding the vertices and asymptotes to graph the hyperbola
Quote:

Originally Posted by Ashz
Okay, I think I'm tracking ~

y=+-(1/6)x
Vertices (0, 1/6) & (0, -1/6)
Foci (0, sqrt 37/36) & (0, -sqrt 37/36)

Be sure to use parentheses! Many people would interpret "sqrt 37/36" to mean sqrt(37)/36 while you mean sqrt(37/36). Of course, that is the same as sqrt(37)/6.
• Nov 27th 2012, 01:22 PM
Ashz
Re: 36y^2-x^2=1 Finding the vertices and asymptotes to graph the hyperbola
Thank you