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Question
Find the equation of the tangent and the normal to the following curve at the indicated point \[y^2 = \frac{x^3}{4 - x}at \left( 2, - 2 \right)\] ?
Solution
\[y^2 =\frac{x^3}{4 - x}\]
\[\text { Differentiating both sides w.r.t.x}, \]
\[2y \frac{dy}{dx} = \frac{\left( 4 - x \right)\left( 3 x^2 \right) - x^3 \left( - 1 \right)}{\left( 4 - x \right)^2} = \frac{12 x^2 - 3 x^3 + x^3}{\left( 4 - x \right)^2} = \frac{12 x^2 - 2 x^3}{\left( 4 - x \right)^2}\]
\[\frac{dy}{dx} = \frac{12 x^2 - 2 x^3}{2y \left( 4 - x \right)^2}\]
\[\text { Given } \left( x_1 , y_1 \right) = \left( 2, - 2 \right)\]
\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( 2, - 2 \right) =\frac{48 - 16}{- 16}=-2\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m\left( x - x_1 \right)\]
\[ \Rightarrow y + 2 = - 2 \left( x - 2 \right)\]
\[ \Rightarrow y + 2 = - 2x + 4\]
\[ \Rightarrow 2x + y - 2 = 0\]
\[\text { Equation of normal is },\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y + 2 = \frac{1}{2} \left( x - 2 \right)\]
\[ \Rightarrow 2y + 4 = x - 2\]
\[ \Rightarrow x - 2y - 6 = 0\]
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