Asked by raman
find the eqyuation of hyperbola traced by points that moves so the difference to (0,0) and (11,11) is 11
Answers
Answered by
oobleck
just plug in your specification:
√(x^2+y^2) = √((x-11)^2 + (y-11)^2) + 11
square both sides to get
x^2+y^2 = (x-11)^2 + (y-11)^2 + 121 + 22√((x-11)^2 + (y-11)^2)
collect terms to get and divide by 11 to get
2x+2y-33 = 2√((x-11)^2 + (y-11)^2)
square again and collect terms and you wind up with
44x + 44y - 8xy = 121
which is, as expected, a rotated hyperbola.
You could have started out by specifying that the foci were at ±11/√2 and the shifted and rotated it, but that would have been a bit more trouble.
√(x^2+y^2) = √((x-11)^2 + (y-11)^2) + 11
square both sides to get
x^2+y^2 = (x-11)^2 + (y-11)^2 + 121 + 22√((x-11)^2 + (y-11)^2)
collect terms to get and divide by 11 to get
2x+2y-33 = 2√((x-11)^2 + (y-11)^2)
square again and collect terms and you wind up with
44x + 44y - 8xy = 121
which is, as expected, a rotated hyperbola.
You could have started out by specifying that the foci were at ±11/√2 and the shifted and rotated it, but that would have been a bit more trouble.
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