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Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 4  Maths Textbook Solution.

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ANSWER: Equation of tangent, y-b \tan t=\left[\frac{b \operatorname{cosect}}{a}\right][x-a \sec t]

                 Equation  of normal , y-b \tan t=\left[\frac{-a \operatorname{\sin t}}{b}\right][x-a \sec t]


 Differentiating  with respect to x  to get its slope.


x=asect,y=b \ tan t\; \; \ \ at\: t


Upon differentiation

\begin{aligned} &\frac{d x}{d t}=a \sec t \tan t, \frac{d y}{d t}=b \sec ^{2} t \\ &\therefore \frac{d y}{d x}=\frac{b \text { cosect }}{a} \\ &M(\text { tangent) at } \end{aligned}

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

m\left ( normal \right )at\: t=\left ( -\frac{3}{6} \right )\sin t

The equation of tangent is given by,

y-b \tan t=\frac{b \operatorname{cossect}}{a}(x-a \sec t)

The equation of Normal  is given by  ,

y-b \tan t=\frac{-a \operatorname{\sin t}}{b}[x-a \sec t]


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