1 Attachment(s)
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 ^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 (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
