Question

Find the point on the curve \( y=\sqrt{x-3} \) where the tangent is perpendicular to the line given by \( 6 x+3 y-5=0 \).

Answer 1

Let the required point on the curve \(y = \sqrt{x - 3} \) be P(x₁, y₁).

Differentiating \(y = \sqrt{x - 3} \) w.r.t.x., we get

∴ Slope of the tangent at (x₁, y₁)

\(= (\frac{dy}{dx})_{at(x_1, y_1)}\)

\(= \frac{1}{2\sqrt{x_1-3}}\)

Since, this tangent is perpendicular to 6x + 3y – 5 = 0 whose slope is -6/3 = –2.

Since, (x₁, y₁) lies on y = √x - 3, y₁ = √x₁ - 3

When x₁ = 4, y₁ = √4 - 3 = ±1

Hence, the required points are (4, 1) and (4, –1).