#### Question

Find the equation of the tangent and the normal to the following curve at the indicated point xy = c^{2} at \[\left( ct, \frac{c}{t} \right)\] ?

#### Solution

\[{xy=c}^2 \]

\[\text { Differentiating both sides w.r.t.x }, \]

\[x\frac{dy}{dx} + y = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- y}{x}\]

\[\text { Given } \left( x_1 , y_1 \right) = \left( ct, \frac{c}{t} \right)\]

\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( ct, \frac{c}{t} \right) =\frac{- \frac{c}{t}}{ct}=\frac{- 1}{t^2}\]

\[\text { Equation of tangent is },\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - \frac{c}{t} = \frac{- 1}{t^2} \left( x - ct \right)\]

\[ \Rightarrow \frac{yt - c}{t} = \frac{- 1}{t^2} \left( x - ct \right)\]

\[ \Rightarrow y t^2 - ct = - x + ct\]

\[ \Rightarrow x + y t^2 = 2ct\]

\[\text{ Equation of normal is },\]

\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]

\[ \Rightarrow y - \frac{c}{t} = t^2 \left( x - ct \right)\]

\[ \Rightarrow yt - c = t^3 x - c t^4 \]

\[ \Rightarrow x t^3 - yt = c t^4 - c\]