# Generate Hyperbola Equation

awaludin1976 shared this question 3 years ago

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. 1

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 1

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? 1

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? 1

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.