Question

Find the equation of the tangent and the normal to the following curve at the indicated point \[c^2 \left( x^2 + y^2 \right) = x^2 y^2 \text { at }\left( \frac{c}{\cos\theta}, \frac{c}{\sin\theta} \right)\] ?

Answer 1

\[c^2 \left( x^2 + y^2 \right) = x^2 y^2 \]

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

\[ \Rightarrow 2x c^2 + 2y c^2 \frac{dy}{dx} = x^2 2y \frac{dy}{dx} + 2x y^2 \]

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

\[ \Rightarrow \frac{dy}{dx} = \frac{x y^2 - x c^2}{y c^2 - x^2 y}\]

\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( \frac{c}{\cos \theta}, \frac{c}{\sin \theta} \right) \]

\[=\frac{\frac{c^3}{\cos \theta \sin^2 \theta} - \frac{c^3}{\cos \theta}}{\frac{c^3}{\sin\theta} - \frac{c^3}{\cos^2 \theta \sin\theta}}\]

\[ = \frac{\frac{1 - \sin^2 \theta}{\cos\theta \sin^2 \theta}}{\frac{\cos^2 \theta - 1}{\cos^2 \theta \sin\theta}}\]

\[ = \frac{co s^2 \theta}{\cos \theta \sin^2 \theta} \times \frac{\cos^2 \theta \sin\theta}{- \sin^2 \theta}\]

\[ = \frac{- \cos^3 \theta}{\sin^3 \theta}\]

\[\text { Given }\left( x_1 , y_1 \right) = \left( \frac{c}{\cos \theta}, \frac{c}{\sin \theta} \right)\]

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

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

\[ \Rightarrow y - \frac{c}{\sin \theta} = \frac{- \cos^3 \theta}{\sin^3 \theta} \left( x - \frac{c}{\cos \theta} \right)\]

\[ \Rightarrow \frac{y\sin\theta - c}{\sin\theta} = \frac{- \cos^3 \theta}{\sin^3 \theta}\left( \frac{x \cos\theta - c}{\cos\theta} \right)\]

\[ \Rightarrow \sin^2 \theta\left( y \sin\theta - c \right) = - \cos^2 \theta\left( x\cos\theta - c \right)\]

\[ \Rightarrow y \sin^3 \theta - c \sin^2 \theta = - x \cos^3 \theta + c \cos^2 \theta\]

\[ \Rightarrow x \cos^3 \theta + y \sin^3 \theta = c\left( si n^2 \theta + \cos^2 \theta \right)\]

\[ \Rightarrow x \cos^3 \theta + y \sin^3 \theta = c\]

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

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

\[ \Rightarrow y - \frac{c}{\sin \theta} = \frac{\sin^3 \theta}{\cos^3 \theta}\left( x - \frac{c}{\cos \theta} \right)\]

\[ \Rightarrow \cos^3 \theta\left( y - \frac{c}{\sin \theta} \right) = \sin^3 \theta\left( x - \frac{c}{\cos \theta} \right)\]

\[ \Rightarrow y \cos^3 \theta - \frac{c \cos^3 \theta}{\sin\theta} = x \sin^3 \theta - \frac{c \sin^3 \theta}{\cos\theta}\]

\[ \Rightarrow x \sin^3 \theta - y \cos^3 \theta = \frac{c \sin^3 \theta}{\cos\theta} - \frac{c \cos^3 \theta}{\sin\theta}\]

\[ \Rightarrow x \sin^3 \theta - y \cos^3 \theta = c\left( \frac{\sin^4 \theta - \cos^4 \theta}{\cos\theta \sin\theta} \right)\]

\[ \Rightarrow x \sin^3 \theta - y \cos^3 \theta = c\left[ \frac{\left( \sin^2 \theta + \cos^2 \theta \right)\left( \sin^2 \theta - \cos^2 \theta \right)}{\cos\theta \sin\theta} \right]\]

\[ \Rightarrow \sin^3 \theta - y \cos^3 \theta =\]

\[ 2c \left[ \frac{- \left( \cos^2 \theta - \sin^2 \theta \right)}{2\cos\theta \sin\theta} \right]\]

\[ \Rightarrow \sin^3 \theta - y \cos^3 \theta = 2c\left[ \frac{- \cos \left( 2\theta \right)}{\sin\left( 2\theta \right)} \right]\]

\[ \Rightarrow \sin^3 \theta - y \cos^3 \theta = - 2c \text { cot }\left( 2\theta \right)\]

\[ \Rightarrow \sin^3 \theta - y \cos^3 \theta + 2c \text { cot }\left( 2\theta \right) = 0\]