Question

Find the equations of the tangent to the curve x^2/3 + y^2/3 = 2 at (1, 1).
1. x + y + 2 = 0 
2. x - y - 2 = 0 
3. 2x + y - 2 = 0 
4. x + y = 0 
5. x + y - 2 = 0 

Answer 1

Correct Answer - Option 5 : x + y - 2 = 0 

Concept:

The slope of the tangent at (x, y) is (m),

m = \(\left( \dfrac{dy}{dx} \right)_{(x,~y)}\)

And,

The equation of the tangent at (x₁, y₁) is,

⇒ (y - y₁) = m (x - x₁)

Given:

x^2/3 + y^2/3 = 2

Calculation:

Differentiating with respect to x,

\(\dfrac{2}{3} x^{-1/3}+ \dfrac{2}{3} y^{-1/3}\dfrac{dy}{dx}=0 \)

\(\dfrac{dy}{dx} = - \left( \dfrac{y}{x}\right)^{1/3}\)

Then,

The slope of the tangent at (1, 1) is,

⇒ \(\left( \dfrac{dy}{dx} \right)_{(x,~y)}\)

⇒ \(\left( \dfrac{dy}{dx} \right)_{(1, 1)}= -1\)

And,

The equation of the tangent at (x1, y1) is,

⇒ (y - y1) = m (x - x1)

So,

The equation of the tangent at (1, 1) is

⇒ (y - 1) = (-1)(x - 1)

⇒ x + y - 2 = 0