How do I find the equation of a hyperbola given a point and one asymptote?

[ZeHgS]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

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 going to 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(b² * (x²/a² - 1)) which I got from the reduced equation of a hyperbola to y = bx/a + d). I ended up with (b² - b)*x² + (2dba)*x + (d²a² + a²b²). Then I supposed B² - 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.

Please help I'm almost committing suicide over these two.

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

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 going to 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(b² * (x²/a² - 1)) which I got from the reduced equation of a hyperbola to y = bx/a + d). I ended up with (b² - b)*x² + (2dba)*x + (d²a² + a²b²). Then I supposed B² - 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.

Please help I'm almost committing suicide over these two.

(a) Try an equation of the form y^2 - a*x^2 = b, where a,b > 0. The {y<0} branch of the hyperbola is y = -sqrt(1 + a*x^2), from which you can get the asymptotic form y =approx= c*x for large x (for some 'c' you need to determine), and match it with the asymptote 5y + 2x = 0. So, now you have one of the parameters 'a' or 'b'. Use this in the equation of the {y > 0} branch, y = +sqrt(b + a*x^2), and use P(6,2) information to get the other parameter.

(b) Suppose the asymptotes run southwest-northeast and southeast-northwest, and suppose the two branches of the hyperbola are in the north and south sectors. Look at the northern branch y = f(x), and note that the slope of the tangent line, f'(x), increases strictly as x increases; the limiting slope as x --> infinity is the asymptote's slope. With this hint you may proceed further.

RGV

 

[ZeHgS]

Got it, thank you!