Find the equation of a curve passing through (2, 1) if the slope of the tangent to the curve at any point (x, y) is

Slope of tangent is given as


Slope of tangent of a curve y = f(x) is given by



Put y = vx



Differentiate vx using product rule,







Integrate



Put 1 – v2 = t hence differentiating with respect to v we get which means 2vdv = -dt




Resubstitute t



Resubstitute v



Now it is given that the curve is passing through (2, 1)


Hence (2, 1) will satisfy the curve equation (a)


Putting values x = 2 and y = 1 in (a)







Using log a + log b = loga b



Put c in equation (a)






Using log a – log b = log




Using log a + log b = log ab




3x = 2(x2 – y2)


3x = 2x2 – 2y2


2y2 = 2x2 – 3x



Hence equation of curve is


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