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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 19 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 19 Maths Textbook Solution.

Answers (1)

Answer: 3y=2 \sqrt{2x}-2

Hint: Differentiate to find the slop of the tangent

Given: x=\sin \: 3t,y=\cos t2t at t=\frac{\pi }{4}

Solution: Slope of tangent is \frac{dy}{dx}

\begin{aligned} &\Rightarrow \frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{d(\cos 2 t)}{d t}}{\frac{d(\sin 3 t)}{d t}}=\frac{-2 \sin 2 t}{3 \cos 3 t} \\ &{\left[\frac{d y}{d x}\right]_{x t-\frac{\pi}{4}}=\frac{-2 \sin \frac{\pi}{2}}{3 \cos \frac{3 \pi}{4}}=\frac{-2 \times 1}{3 \times\left[-\frac{1}{\sqrt{2}}\right]}=\frac{2 \sqrt{2}}{3}} \end{aligned}

Now,x=\sin \left [ \frac{3\pi }{4} \right ]=\frac{1}{\sqrt{2}}

y=\cos \left [ \frac{2\pi }{4} \right ]=0

\therefore Equation of the tangent is

\begin{aligned} &y-0=\frac{d y}{d x}\left[x-\left[\frac{1}{\sqrt{2}}\right]\right] \\ &y=\frac{2 \sqrt{2}}{3}\left[x-\frac{1}{\sqrt{2}}\right] \\ &y=\frac{2 \sqrt{2}}{3} x-\frac{2}{3} \end{aligned}

or 3y=2\sqrt{2x}-2

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