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 [itex]49x^2- 49y^2= 441[/itex] is wrong.

As x and y get very large, "441" will be very very very small in comparison with [itex]49x^2[/itex] and [itex]49y^2[/itex] and can be neglected. That is the graph will be very close to the graph of [itex]49x^2- 49y^2= 0[/itex] which is the same as [itex]x^2- y^2= (x- y)(x+ y)= 0[/math] so y= -x and y= x are the asymptotes.

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?

 

[Architeuthis Dux]

To derive the equation for the asymptote y=7/3x, we need to understand the definition of an asymptote. An asymptote is a line that a curve approaches but never touches. In the case of a hyperbola, the asymptote is the line that the hyperbola's branches approach as they extend towards infinity.

To find the equation of the asymptote, we can use the general equation of a hyperbola: (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where (h,k) is the center of the hyperbola, and a and b are the distance from the center to the vertices in the x and y directions respectively.

In this case, the center of the hyperbola is (0,0) since the vertices are at (3,0) and (-3,0). So, plugging in the values, we get (x-0)^2/3^2 - (y-0)^2/b^2 = 1.

Now, since the asymptote is y=7/3x, we can substitute this equation into the hyperbola equation to solve for b. This gives us (x-0)^2/3^2 - (7/3x-0)^2/b^2 = 1. Simplifying, we get 9x^2 - 49x^2/b^2 = 1.

To find b, we can use the fact that the distance from the center to the vertices is a and b in the x and y directions respectively. So, the distance between the vertices is 2a, which in this case is 6. We also know that the slope of the asymptote is 7/3, which is equal to b/a. So, b/a = 7/3, and solving for b, we get b=7a/3.

Plugging this value of b into our equation, we get 9x^2 - 49x^2/(7a/3)^2 = 1. Simplifying, we get 9x^2 - 9x^2/a^2 = 1. Solving for a^2, we get a^2 = 9, which means a=3.

Now, plugging in the values of a and b into our hyperbola equation, we get (x-0)^2/