9euraka why shudnt we use hyperbolic eq
For any real t ,x=2+[(et+e-t)/2],
y=2+[(et-e-t)/2]
is a point on hyperbola x2-y2-4x+4y-1=0.Find area bounded by hyperbola and the lines joining the centre to the points corresponding to t1 and -t1
(Plz dont solve it using hyperbolic functions)
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2 Answers
62x=2+[(et+e-t)/2]
dx/dt = [(et-e-t)/2] = y - 2
similarly
dy/dt = [(et+e-t)/2] = x - 2
hyperbola is (x-2)2-(y-2)2=1
if u substitute x-2= X and y-2=Y
then u will reduce the whole thing to X2-Y2=1
and the area with the center to the points corresponding to t1 and t2
where new X and Y corespondingly will be [(et+e-t)/2] and [(et-e-t)/2]
