# Find the equation of the hyperbola with centre at the origin

1. Mar 7, 2005

### blue_soda025

Find the equation of the hyperbola with centre at the origin and sketch the graph.
e. tranverse axis is on the y-axis and passes through the points R(4, 6) and S(1, -3)

How would I find a and b? I plugged in the coordinates in $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ and came up with two equations. The thing is, I get to the point where I have to take the square root of both sides, but one side is always negative. I thought you couldn't take the square root of a negative?

2. Mar 8, 2005

### dextercioby

What are the 2 equations...?Ang why is one part always negative...?

Daniel.

3. Mar 8, 2005

### HallsofIvy

Have you considered that
$$\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1$$
is also an equation of a hyperbola?

4. Mar 8, 2005

### blue_soda025

The equations are:

$$b = \pm\sqrt{\frac{36a^2}{16 - a^2}}$$
$$b = \pm\sqrt{\frac{9a^2}{1 - a^2}}$$

Then when I try to solve for a after combining them, one side is always negative. I don't know why it is always negative, but I've tried solving for b instead of a first, and I encounter the same problem. I think someone else in my class had the same problem too. Maybe I just did it wrong or something..
The teacher said something about subtracting those two equations once you get them..? How would you do that?

5. Mar 8, 2005

### HallsofIvy

Well, you just subtract! $0= \sqrt{\frac{36a^2}{16-a^2}}- \sqrt{\frac{9a^2}{1-a^2}}$ and now "move" that second term to the other side of equation: $\sqrt{\frac{36a^2}{16-a^2}}= \sqrt{\frac{9a^2}{1-a^2}}$ and now square both sides.

Do you see where those formulas are from? If you assume a formula of the form $\frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$, then putting x= 4, y= 6 gives $\frac{16}{a^2}- \frac{36}{b^2}= 1$ and putting x= 1, y= -3 gives $\frac{1}{a^2}-\frac{9}{b^2}= 1$. Solve each of those for b and you get those two equations. Actually, what I would do is Multiply that second equation by 4 to bet $\frac{4}{a^3}-\frac{36}{b^2}= 4$. Now subtract the first equation from that to eliminate b: $\frac{12}{a^2}= -3$ . Yes! You do have a square on one side equal to a negative number!

Okay, so do what I suggested before: try writing your basic equation as
$$\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1$$
and do the same as before.

6. Mar 8, 2005

### blue_soda025

Oh, I think I finally got the answer. I should've used $$\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1$$ since the graph opens up and down. I could've sworn I did that when I attempted this the first time, but I couldn't get the answer for some reason.. Well, nevermind that. Thanks a lot for your help!