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Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 7 Maths Textbook Solution.

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ANSWER: The equation of normal y-a m^{3}=\left(\frac{-2}{3 m}\right)\left(x-a m^{2}\right)


Differentiating  with respect to  x to get its slope of tangent  2ay\left ( \frac{dy}{dx} \right )=3x^{2}


ay^{2}=x^{3}at\; the \: point\left ( am^{2},am^{2} \right )


Upon differentiation


m(tangent) at \left ( am^{2},am^{3} \right )is \frac{3m}{2}

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

m(tangent) at at \left ( am^{2},am^{3} \right )\: is \frac{-2}{3m}

The equation of Normal  is given by  ,

\Rightarrow y-a m^{3}=\left(-\frac{2}{3 m}\right)\left(x-a m^{2}\right)

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