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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.

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

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


 Differentiating  with respect to x  to get its slope.

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


Upon differentiation


\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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