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  • Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 3 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 3 Maths Textbook Solution.

Answers (1)

Answer:: Equation of tangent, y-2a=1\left ( x-a \right )

              Equation  of normal , y-2a=1\left ( x-a \right )

Hint:

 Differentiating  with respect to x  to get its slope.

Given: x-at^{2},y=4at\: \: at \: \: t=1

Solution:

Upon differentiation

\frac{dx}{dt}=2at,\frac{dy}{dt}=2a

\therefore \frac{dy}{dt}=\frac{1}{t}

M(tangent) at t=1 is 1

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

m\left (Normal \right )at\: t=1\: is \; -1

The equation of tangent is given by,

y-y_{1}=m\left ( tangent \right )\left [ x-x_{1} \right ]

\Rightarrow y-2a=1\left ( x-a \right )

The equation of Normal  is given by  ,

y-y_{1}=m\left ( tangent \right )\left [ x-x_{1} \right ]

\Rightarrow y-2a=-1\left ( x-a \right )

 

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