Question

The equation of tangent to the curve y(1 + x²) = 2 – x, where it crosses x-axis is ______.

Options

• x + 5y = 2

• x – 5y = 2

• 5x – y = 2

• 5x + y = 2

Answer 1

The equation of tangent to the curve y(1 + x²) = 2 – x, where it crosses x-axis is x + 5y = 2.

Explanation:

Given that y(1 + x²) = 2 – x    ...(i)

If it cuts x-axis, then y-coordinate is 0.

∴ 0(1 + x²) = 2 – x

⇒ x = 2

Put x = 2 in equation (i)

y(1 + 4) = 2 – 2

⇒ y(5) = 0

⇒ y = 0

Point of contact = (2, 0)

Differentiating equation (i) w.r.t. x, we have

`y xx 2x + (1 + x^2)  "dy"/"dx"` = – 1

⇒ `2xy + (1 + x^2) "dy"/"dx"` = – 1

⇒ `(1 + x^2) "dy"/"dx"` = – 1 – 2xy

∴ `"dy"/"dx" = (-(1 + 2xy))/((1 + x^2))`

⇒ `"dy"/"dx"_(2, 0) = (-1)/((1 + 4)) = (-1)/5`

Equation of tangent is y – 0 = `- 1/5 (x - 2)`

⇒ 5y = – x + 2

⇒ x + 5y = 2