Kenneth S. answered • 07/24/17
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First, sketch it knowing that center is at origin and y intercepts are ±6.
This means that equation is of form y2/a2 - x2/b2 = 1.
Note that the transverse axis is the y axis, and distance from center(origin) to foci is c=6.
The slope of one asymptote is a/b = 1/3 and this gives us b2 = 9a2.
Since for any hyperbola c2 -a2 = b2 we find that 36 = 10a2 so a=6/√10 and b=324/√10.
Now you can substitute for a squared & b squared, and simplify!
Kenneth S.
If you know how the general equation of an hyperbola is obtained, i.e. the proof or derivation of the equation, you'll realize that the center is chosen as the origin and (in this case) the transverse axis runs along the y-axis. Along this line, one proceeds a distance c to the foci (one in each direction). Along this line,one proceeds a distance of a, a<c, to points that will be the y-intercepts of the hyperbola's two branches.
Then the distance formula was used to express the defining property of this hyperbola, that from any point P on the hyperbola: PF1 - PF2 = 2a. The algebra leads to a point where b2 is introduced as a surrogate for c2-a2.
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