# Find equation of a Hyperbola that passes through two ordered pairs

1. Mar 26, 2009

### synergix

1. The problem statement, all variables and given/known data

Find an equation of the hyperbola that passes through the points (-3,-2) and (4, sqrt(5))

2. Relevant equations

x^2/a^2 -y^2/b^2=1 or y^2/a^2 - x^2/b^2 = 1

3. The attempt at a solution
To solve this problem I first started by setting up two equations containing the two ordered pairs. Then I solved for b for (1) and then substituted that into (2).

I used the standard form X^2/a^2 - y^2/b^2 = 1 for both

so..

(1) 9/a^2 - 4/b^2 = 1 and (2) 16/a^2 - 5/b^2 = 1

and...
I didn't get answers that checked.
I am wondering if I am even allowed to do that substitution I don't know if that is correct or not so what I am trying to determine is whether I am approaching this problem incorrectly or if my approach is sound but a mistake was made somewhere in my calculations?

2. Mar 26, 2009

### HallsofIvy

Staff Emeritus
First, just as there are an infinite number of circles that pass through two given points, so there are an infinite number of hyperbolas that pass through two given points. You are also assuming that the axes of symmetry are the x and y axes and that the center of the hyperbola is the origin though you do not give those as conditions.

Certainly, assuming that, the equation can be written as $x^2/a^2- y^2/b^2= 1$ or $y^2/b^2- x^2/a^2= 1$. Trying the first, yes, the fact that the hyperbola goes through (-3, -2) gives $9/a^2- 4/b^2= 1$ and the fact that it goes through $(4,\sqrt{5})$ gives $16/a^2- 5/b^2= 1$.

But it is impossible to say what you are doing wrong if you don't show what you did. HOW did you solve those equations and what result did you get? If you did not get a correct result, did you then try $y^2/b^2- x^2/a^2= 1$?

3. Mar 26, 2009

### synergix

Ok I made a mistake... I have tried again using Y^2/a^2-x^2/b^2=1

so I am solving the first equation for a
To eliminate the fractions I times both sides by a squared times b squared
(4/a^2-9/b^2=1)a^2b^2

4b^2-9a^2-a^2b^2=0

now I am stuck :( BTW my first attempt was incorrect(obviously) but I only got passed this point because I incorrectly manipulated the equations. I am not sure how to isolate the "a" variable here I will try to figure it out on my own in the meantime but if you could help that would be great.

4. Mar 26, 2009

### synergix

Alright I managed it. I started this problem late last night(lame excuse) and I made a mistake when I was rearranging the equations but anyways I have solved it now and my answers checked thx!

Last edited: Mar 26, 2009