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  • Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 5 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 5  Maths Textbook Solution.

Answers (1)

ANSWER: Equation of tangent, \begin{aligned}& y-a[1-\cos \theta]=\left[\frac{\sin \theta}{1+\cos \theta}\right] (x-a(\theta+\sin \theta)) \end{aligned}

                 Equation  of normal, \begin{aligned}y-a[1-\cos \theta]=-\left[\frac{1+\cos \theta}{\sin \theta}\right] (x-a(\theta+\sin \theta) )\end{aligned}

HINTS:

 Differentiating  with respect to  \theta to get its slope.

GIVEN:

x=a\left ( \theta +\sin \theta \right ) \ \ \ y=a\left ( 1-\cos \theta \right )at\; \theta

SOLUTION:

Upon differentiation

\frac{dx}{d\theta }=a\left ( 1+\cos \theta \right ),\frac{dy}{d\theta }=a\sin \theta

\therefore \frac{dy}{dx}=\frac{\sin \theta }{1+\cos \theta }

\\\text{m (tangent) at } \theta \ is = \frac{\sin \theta }{1+\cos \theta }

The normal is perpendicular to tangent, therefore , m_{1}m_{2}=-1

m\left ( normal \right )at\: \theta \ is \ \ =\frac{-(1+\cos \theta) }{\sin \theta }

The equation of tangent is given by,

\begin{aligned} &y-y_{1}=m(\text { tangent })\left(x-x_{1}\right) \\ &\Rightarrow y-a[1-\cos \theta]=\left[\frac{\sin \theta}{1+\cos \theta}\right] (x-a(\theta+\sin \theta)) \end{aligned}

The equation of Normal  is given by  ,

\begin{aligned} &y-y_{1}=m(\text { normal })\left(x-x_{1}\right) \\ &\Rightarrow y-a[1-\cos \theta]=-\left[\frac{1+\cos \theta}{\sin \theta}\right] (x-a(\theta+\sin \theta) )\end{aligned}

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