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