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प्रश्न
Find the equation of the tangent and the normal to the following curve at the indicated point 4x2 + 9y2 = 36 at (3cosθ, 2sinθ) ?
उत्तर
\[4 x^2 + 9 y^2 = 36\]
\[\text { Differentiating both sides w.r.t.x }, \]
\[8x + 18y \frac{dy}{dx} = 0\]
\[ \Rightarrow 18y \frac{dy}{dx} = - 8x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 8x}{18y} = \frac{- 4x}{9y}\]
\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_\left( 3 \cos\theta, 2 \sin\theta \right) =\frac{- 12\cos\theta}{18\sin\theta}=\frac{- 2 \cos\theta}{3 \sin\theta}\]
\[\text { Given} \left( x_1 , y_1 \right) = \left( 3 \cos\theta, 2 \sin\theta \right)\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 2 \sin\theta = \frac{- 2 \cos\theta}{3 \sin\theta}\left( x - 3 \cos\theta \right)\]
\[ \Rightarrow 3y \sin\theta - 6 \sin^2 \theta = - 2x \cos\theta + 6 \cos^2 \theta\]
\[ \Rightarrow 2x \cos\theta + 3y \sin\theta = 6\left( \cos^2 \theta + \sin^2 \theta \right)\]
\[ \Rightarrow 2x \cos\theta + 3y \sin\theta = 6\]
\[\text { Equation of normal is },\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - 2 \sin\theta = \frac{3 \sin\theta}{2 \cos\theta}\left( x - 3 \cos\theta \right)\]
\[ \Rightarrow 2y \cos\theta - 4 \sin\theta \cos\theta = 3x \sin\theta - 9 \sin\theta \cos\theta\]
\[ \Rightarrow 3x \sin\theta - 2y \cos\theta - 5\sin\theta \cos\theta = 0\]
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