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Question
Find the equation of the tangent and the normal to the following curve at the indicated point \[x^\frac{2}{3} + y^\frac{2}{3}\] = 2 at (1, 1) ?
Solution
\[x^\frac{2}{3} + y^\frac{2}{3} = 2\]
\[\text { Differentiating both sides w.r.t.x}, \]
\[\frac{2}{3} x^\frac{- 1}{3} + \frac{2}{3} y^\frac{- 1}{3} \frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- x^\frac{- 1}{3}}{y^\frac{- 1}{3}} = \frac{- y^\frac{1}{3}}{x^\frac{1}{3}}\]
\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) =\frac{- 1}{1}=-1\]
\[\text { Given } \left( x_1 , y_1 \right) = \left( 1, 1 \right)\]
\[\text { Equation of tangent is,}\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 1 = - 1\left( x - 1 \right)\]
\[ \Rightarrow y - 1 = - x + 1\]
\[ \Rightarrow x + y - 2 = 0\]
\[\text { Equation of normal is },\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - 1 = 1\left( x - 1 \right)\]
\[ \Rightarrow y - 1 = x - 1\]
\[ \Rightarrow y - x = 0\]
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