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

Explanation

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

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

`\frac{2}{3} x^{-\frac{1}{3}}+\frac{2}{3} y^{-\frac{1}{3}} \frac{d y}{d x}=0 `

`\Rightarrow y^{\frac{1}{3}}+x^{\frac{1}{3}} \frac{d y}{d x}=0 `

`\Rightarrow \frac{d y}{d x}=-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}} \cdot(\because \frac{d}{d x} x^{n}=n x^{n-1})`

Therefore, the slope of tangent to the given curve `x^{\frac{2}{3}}+y^{\frac{2}{3}}=2 is \frac{d y}{d x}=-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}`

The slope of normal to the given curve is `\frac{1}{\frac{d y}{d x}}=\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.)

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