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Generate Hyperbola Equation
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Dear All
I see that Geogebra is able to generate hyperbola equation when it is given the foci and one point in the hyperbola. I wonder how geogebra do this? especially when the foci are not located at the axes. Thank you.
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If you type this in the CAS View, it will give you the (very long) general equation:
Hyperbola[(a,b),(c, d),(p,q)]
If you also define
A=(1,2)
B=(3,4)
C=(1,4)
a=x(A)
b=y(A)
c=x(B)
d=y(B)
p=x(C)
q=y(C)
then you can type the long formula into the Input Bar to check it works:
https://tube.geogebra.org/m/2399301
Thank you very much. I think I don't have to question the long equation :)
Firstly I thought it was an exact equation, but sometime when I gave the value of foci and point the graph doesn't always go through the point. Now I can assume that it is an approximate equation not an exact equation, correct?
The equation is exact, but the drawing algorithm is numeric so may not always work. What are the coordinates of A, B, C when it fails?
Hyperbola[(1, 0), (0, 1), (2, 10000)]
Hyperbola[(1, 0), (1,0), (2, 10000)]
I expect to see intersection at (2,10000) but in my computer the hyperbolas doesn't have it at the expected value.
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