Question

Find the equation of tangent to the curve y = `sqrt(x - 3)` which is perpendicular to the line 6x + 3y – 4 = 0

Answer 1

Equation of the curve is y = `sqrt(x - 3)`

Differentiating w.r.t. x, we get

`"dy"/"dx" = 1/(2sqrt("x - 3"))`

Slope of the tangent at P(x₁, y₁) is

`(("d"y)/("d"x))_((x_1","  y_1)) = 1/(2sqrt(x_1 - 3))`

Slope of the line 6x + 3y – 4 = 0 is `(-6)/2` = – 2.

According to the given condition, tangent to the curve is perpendicular to the line 6x + 3y – 4 = 0.

∴ slope of the tangent = `(("d"y)/("d"x))_((x_1, y_1)` = `1/2`

∴ `1/(2sqrt(x_1 - 3)) = 1/2`

∴ `sqrt(x_1 - 3) = 1`

∴ x₁ – 3 = 1

∴ x₁ = 4

P(x₁, y₁) lies on the curve y = `sqrt(x - 3)`

∴ `y_1 = sqrt(4 - 3)`

∴ y₁ = 1

∴ The point on the given curve is (4, 1).

∴ Equation of the tangent at (4, 1) is

(y – 1) = `1/2` (x – 4)

∴ 2y – 2 = x – 4

∴ x – 2y – 2 = 0