Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve any point (x, y) is

Slope of tangent is given as


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




Integrate



Using partial fraction for




Equating numerator


A(x + 1) + Bx = 1


Put x = 0


A = 1


Put x = -1


B = -1


Hence


Hence equation (a) becomes




log(y – 1) = logx – log(x + 1) + c …(b)


Now it is given that the curve is passing through (1, 0)


Hence (1, 0) will satisfy the curve equation (b)


Putting values x = 1 and y = 0 in (b)


If we put y = 0 in (b) we get log (-1) which is not defined hence we must simplify further equation (b)


log (y – 1) – log x = – log (x + 1) + c


Using log a – log b = log




Using log a + log b = log ab



Take the constant c as log c so that we don’t have any undefined terms in our equation (Why only log c and not any other term because taking log c completely eliminates the log terms so we don’t have to worry about the undefined terms appearing in our equation)



Eliminating log



Now substitute x = 1 and y = 0



c = -2


Put back c = -2 in (c)



Hence the equation of curve is (y – 1)(x + 1) = -2x


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