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

AI Thread Summary
To find the equation of a hyperbola given a point P(6, 2) and an asymptote line 2x + 5y = 0, one approach is to use the general form of the hyperbola equation and match it with the asymptote to determine parameters. The asymptote's slope can help identify one parameter, while substituting the point P into the equation can yield the other parameter. For the second problem, it is suggested that a line parallel to an asymptote intersects the hyperbola at a single point, which can be proven by analyzing the behavior of the hyperbola's branches and their tangent slopes. The discussion emphasizes the need for a solid understanding of hyperbola properties and equations to solve these problems effectively.
ZeHgS
Messages
2
Reaction score
0

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.
 
Physics news on Phys.org


ZeHgS said:

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
 


Got it, thank you!
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top