​ Find the equations of the tangent and normal to the curve x^(2/3) + y^(2/3) = 2 at (1, 1). ​

Question

Find the equations of the tangent and normal to the curve x^{\(\frac{2}{3}\)} + y^{\(\frac{2}{3}\)} = 2 at (1, 1).

Answer 1

Given curve is x^{\(\frac{2}{3}\)} + y^{\(\frac{2}{3}\)} = 2

Now, differentiate above equation with respect to x, we get

Therefore, the slope of tangent to the given curve   x^{\(\frac{2}{3}\)} + y^{\(\frac{2}{3}\)} = 2 is \(\frac{dy}{dx}\) = -\(\cfrac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}\)

The slope of normal to the given curve is \(\frac{1}{\frac{dy}{dx}}\)= \(\frac{-1}{\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}}\) = \({\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}}}\)

(Because, the multiple of the slope of tangent and the slope of normal to the curve is −1. )

The slope of tangent to the given curve at (1, 1) is \(\bigg[\frac{dy}{dx}\bigg]_{(1,1)}\) is = \(\bigg[-{\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}}}\bigg]_{(1,1)}\) = \({\frac{1^{\frac{1}{3}}}{1^{\frac{1}{3}}}}\) = -1.

And the slope of normal to the given curve at (1, 1) is  

Since, the equation of line passing through (x₁,y₁ ) is given by y - y₁ = \(\bigg[\frac{dy}{dx}\bigg]_{(x_1,y_1)}\) \((x - x_1)\)

The equation of tangent to the given curve at (1, 1) is − 1 = −1(x − 1) ⇒ x + y − 2 = 0.

 The equation of normal to the given curve at (1, 1) is − 1 = 1(x − 1) ⇒ x − y = 0 ⇒ y = x.