Dear Student x=sin 3t and y=cos 2tdxdt=3 cos 3t and dydt=-2 sin 2t∴dydx=dydtdxdt=-2 sin 2t3 cos 3tSlope of tangent, m = dydxt=π4=--2 sin π23 cos 3π4=-2-32=223x1=sin 3×π4=12 and y1=cos 2×π4=0So, x1, y1=12, 0Equation of tangent is,y-y1=m x-x1⇒y-0=223x-12⇒3y=22x-2⇒22x-3y-2=0 Regards

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Dear Student

x = \sin 3 t and y = \cos 2 t \frac{d x}{d t} = 3 \cos 3 t and \frac{d y}{d t} = - 2 \sin 2 t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 2 \sin 2 t}{3 \cos 3 t} Slope of tangent, m = {(\frac{d y}{d x})}_{t = \frac{π}{4}} =- \frac{- 2 \sin (\frac{π}{2})}{3 \cos (\frac{3 π}{4})} = \frac{- 2}{\frac{- 3}{\sqrt{2}}} = \frac{2 \sqrt{2}}{3} x_{1} = \sin (3 \times \frac{π}{4}) = \frac{1}{\sqrt{2}} and y_{1} = \cos (2 \times \frac{π}{4}) = 0 So, (x_{1}, y_{1}) = (\frac{1}{\sqrt{2}}, 0) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 0 = \frac{2 \sqrt{2}}{3} (x - \frac{1}{\sqrt{2}}) \Rightarrow 3 y = 2 \sqrt{2} x - 2 \Rightarrow 2 \sqrt{2} x - 3 y - 2 = 0

Regards

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