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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.

Thanks in advance ~

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

You can write it in the form $\displaystyle \frac{y^2}{\left(\frac{1}{6}\right)^2}- \frac{x^2}{1^2}= 1$. 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 $\displaystyle 36y^2- x^2= (6y- x)(6y+ x)= 0$. That will be true as long as either 6y- x= 0 or 6y+ x= 0 which give straight lines.

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)

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.

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