Question

The equation of curve through the point (1, 0), if the slope of the tangent to t e curve at any point (x, y) is `(y - 1)/(x^2 + x)`, is

Options

• (y + 1)(x – 1) + 2x = 0

• (y + 1)(x – 1) – 2x = 0

• (y – 1)(x – 1) + 2x = 0

• (y – 1)(x + 1) + 2x = 0

Answer 1

(y – 1)(x + 1) + 2x = 0

Explanation:

Here, the slope of the tangent to the curve at any point (x, y) is `(y - 1)/(x^2 + x)`.

∴ `(dy)/(dx) = (y - 1)/(x^2 + x) ⇒ (dy)/(y - 1) = (dx)/(x^2 + x)`

On integrating both sides, we get

`int (dy)/(y - 1) = int (dx)/(x(x - 1))`

⇒ `log (yy - 1) = log x - log (x + 1) + log C`

`log (y - 1) = log ((xC)/(x + 1))`

⇒ `(y - 1)(x + 1)` = xC

Since, the above curve passes through (1, 0)

⇒ `(-1)(2) = 1.C`

⇒ C = – 2

∴ Required equation of the curve is (y – 1)(x + 1) + 2x = 0