Equation of Asymptote (Hyperbola)

[jOANIE]

Homework Statement

What is an equation for the hyperbola with vertices (3,0) and (-3,0) and asymptote y=7/3x?

Homework Equations

The Attempt at a Solution

I solved this problem but still have a question. The answer is 49x^2-49y^2=441 (I solved it by graphing). However, my question: How do I derive the equation for the asymptote y=7/3x?

 

[HallsofIvy]

You can't. If the asymptote is really y= 7/3 x, then $49x^2- 49y^2= 441$ is wrong.

As x and y get very large, "441" will be very very very small in comparison with $49x^2$ and $49y^2$ and can be neglected. That is the graph will be very close to the graph of $49x^2- 49y^2= 0$ which is the same as $x^2- y^2= (x- y)(x+ y)= 0[/math] so y= -x and y= x are the asymptotes.<br /> <br /> If y= (7/3)x is an asymptote and the vertices are (-3,0) and (3,0) then the hyperbola is symmetric about the x-axis and y= (-7/3)x is also an asymptote. Of course y= (7/3)x is the same as 3y= 7x or 7x- 3y= 0. That means that, for large x and y, the graph is close to (7x- 3y)(7x+ 3y)= 49x^2- 9y^2= 0. Can you complete the problem from there?$