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Question
Find the equation of the tangent and the normal to the following curve at the indicated point xy = c2 at \[\left( ct, \frac{c}{t} \right)\] ?
Solution
\[{xy=c}^2 \]
\[\text { Differentiating both sides w.r.t.x }, \]
\[x\frac{dy}{dx} + y = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- y}{x}\]
\[\text { Given } \left( x_1 , y_1 \right) = \left( ct, \frac{c}{t} \right)\]
\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( ct, \frac{c}{t} \right) =\frac{- \frac{c}{t}}{ct}=\frac{- 1}{t^2}\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - \frac{c}{t} = \frac{- 1}{t^2} \left( x - ct \right)\]
\[ \Rightarrow \frac{yt - c}{t} = \frac{- 1}{t^2} \left( x - ct \right)\]
\[ \Rightarrow y t^2 - ct = - x + ct\]
\[ \Rightarrow x + y t^2 = 2ct\]
\[\text{ Equation of normal is },\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - \frac{c}{t} = t^2 \left( x - ct \right)\]
\[ \Rightarrow yt - c = t^3 x - c t^4 \]
\[ \Rightarrow x t^3 - yt = c t^4 - c\]
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