Find the equation of the hyperbola with centre at the origin

In summary, Daniel was trying to solve for a and b in a hyperbola, but one side was always negative. He found that if he assumed a form of the equation \frac{x^2}{a^2}- \frac{y^2}{b^2}= 1, then the equation became simpler. He then solved for b and found that it was \pm\sqrt{\frac{36a^2}{16-a^2}} and \pm\sqrt{\frac{9a^2}{1-a^2}}.
  • #1
blue_soda025
26
0
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 [tex]\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1[/tex] 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? :confused:
 
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  • #2
What are the 2 equations...?Ang why is one part always negative...?

Daniel.
 
  • #3
Have you considered that
[tex]\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1[/tex]
is also an equation of a hyperbola?
 
  • #4
dextercioby said:
What are the 2 equations...?Ang why is one part always negative...?

Daniel.
The equations are:

[tex]b = \pm\sqrt{\frac{36a^2}{16 - a^2}}[/tex]
[tex]b = \pm\sqrt{\frac{9a^2}{1 - a^2}}[/tex]

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

Do you see where those formulas are from? If you assume a formula of the form [itex]\frac{x^2}{a^2}- \frac{y^2}{b^2}= 1[/itex], then putting x= 4, y= 6 gives [itex]\frac{16}{a^2}- \frac{36}{b^2}= 1[/itex] and putting x= 1, y= -3 gives [itex]\frac{1}{a^2}-\frac{9}{b^2}= 1[/itex]. 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 [itex]\frac{4}{a^3}-\frac{36}{b^2}= 4[/itex]. Now subtract the first equation from that to eliminate b: [itex]\frac{12}{a^2}= -3[/itex] . 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
[tex]\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1[/tex]
and do the same as before.
 
  • #6
Oh, I think I finally got the answer. I should've used [tex]\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1[/tex] 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!
 

1. What is a hyperbola?

A hyperbola is a type of conic section that is formed when a plane intersects a double cone at a certain angle. It has two separate curves that are symmetrical about its center.

2. What is the equation of a hyperbola with center at the origin?

The general equation for a hyperbola with center at the origin is (x^2 / a^2) - (y^2 / b^2) = 1, where a and b are the distances from the center to the vertices along the x and y axes, respectively.

3. How do you find the equation of a hyperbola with given information?

To find the equation of a hyperbola, you will need to know the center coordinates, the distance from the center to the vertices, and the type of hyperbola (horizontal or vertical). Using this information, you can plug these values into the general equation and solve for the remaining variables.

4. What is the significance of the center in a hyperbola?

The center of a hyperbola is the point where its two curves intersect. It is also the midpoint between the two foci, which are the two fixed points that determine the shape of the hyperbola.

5. Can a hyperbola have a negative center?

No, a hyperbola cannot have a negative center because the center represents the origin of the coordinate system. However, the center can have negative coordinates, but they are always written as (0,0) for convenience.

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