1. ## Hyperbolas

I'm not sure whether this should go on basic algebra or here, but since I'm learning this in college I'm gonna post it here. Ok, I'm in dire need of help with two problems:

1) A hyperbola goes through the point P(6, 2), and one of its asymptotes is the line r: 2x + 5y = 0. Determine its equation.

2) Prove that a line parallel to one asymptote of a hyperbola interesects it in a single point.

I spent almost two hours on that first problem and tried everything I could, it sounds so fing simple yet I couldn't figure it out. Everything I tried filled up 3-4 pages of my notebook. I must not have understood some basic concept. I'm not gonna transcribe everything here because I think it would be pointless, but please believe me when I said I tried it until I couldn't stand it anymore. If you guys could just point me in the right direction...

For the second one I equaled y = sqrt(b^2 * (x^2/a^2 - 1)) which I got from the reduced equation of a hyperbola to y = bx/a + d). I ended up with (b^2 - b)*x^2 + (2dba)*x + (d^2*a^2 + a^2*b^2). Then I supposed B^2 - 4AC should equal 0, which would mean it had only one solution but I just got a really ugly equation and couldn't see why it would equal 0.

2. For the first part (and probably also for the second part), it will help you to show that if $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is the expression for the hyperbola, then the slopes of the asymptotes are $\pm \frac{b}{a}$, so that the equations of the asymptotes are $y=\pm \frac{b}{a}x$.

Why? Solve the equation for the hyperbola with respect to y to get $y=\pm b\sqrt{\frac{x^2}{a^2}+1}$, and check that as x increases, the right hand side approaches $y=\pm \frac{b}{a}x$.

The slope of your given asymptote is -2/5, so b/a = 2/5. Using the point P(6,2) as well gives you two equations with two unknowns.

3. Originally Posted by HappyJoe
For the first part (and probably also for the second part), it will help you to show that if $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is the expression for the hyperbola, then the slopes of the asymptotes are $\pm \frac{b}{a}$, so that the equations of the asymptotes are $y=\pm \frac{b}{a}x$.
But how do I do that? Would I have to use limits?

Originally Posted by HappyJoe
Why? Solve the equation for the hyperbola with respect to y to get $y=\pm b\sqrt{\frac{x^2}{a^2}+1}$, and check that as x increases, the right hand side approaches $y=\pm \frac{b}{a}x$.
Sorry, I don't get how that would help me.

Originally Posted by HappyJoe
The slope of your given asymptote is -2/5, so b/a = 2/5. Using the point P(6,2) as well gives you two equations with two unknowns.
So I have the equation from the quotation above, and also 2a=5b, right? So plugging in 6 for x and 2 for y on the above and substituting a for 5b/2, I end up getting ${b}^{2 } - 4b + - \frac{144}{25{b}^{2}} + 5 = 0$ but then I'm stuck.

4. Originally Posted by zehgs
But how do I do that? Would I have to use limits?
Yeah.

Originally Posted by zehgs
Sorry, I don't get how that would help me.
How do you mean? It helps you to show that the asymptotes have that particular slope, which will express a in terms of b.

Edit: Oh, and it should be -1 inside of the square root instead of +1.

Originally Posted by zehgs
So I have the equation from the quotation above, and also 2a=5b, right? So plugging in 6 for x and 2 for y on the above and substituting a for 5b/2, I end up getting ${b}^{2 } - 4b + - \frac{144}{25{b}^{2}} + 5 = 0$ but then I'm stuck.
I'm not sure what you did to your equation, but it does look nasty. Substituting 5b/2 for a in $\frac{36}{a^2} - \frac{4}{b^2} = 1$ gives

$\frac{36}{\frac{25}{4}b^2} - \frac{4}{b^2} = 1,$

where I can put the left hand side on the common denominator $\frac{25}{4}b^2$.

5. Yay, got it. Thank you very much.