Differentiation

Question

The equation of tangen to the curve y(1+x²) =2-x where it crossed x-axis is

Answer 1

We have, equation of the curve y(1+x²) = 2−x ...(i)

∴ y.(0+2x) + (1+x²) . dy/dx = 0−1 [on differentiating w.r.t.x]

Since, the given curve passes through x− axis i.e., y = 0.

∴ 0(1+x²) = 2−x [using Eq.(i)]

⇒ x = 2

So, the curve passes through the point (2,0).

∴ (dy/dx)_{(2, 0)} \(=\frac{-1-2\times0}{1+2^2}\) = −1/5 = slope of the curve

∴ slope of tangent to the curve = −1/5

∴ Equation of tangent of the curve passing through (2, 0) is