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  • Provide solution for RD Sharma maths class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 7 maths text book solution.

Provide solution for RD Sharma maths class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 7 maths text book solution.

Answers (1)

Answer : 

             \text { slope of normal }=\frac{1}{t^{2}}

Hint : Using formula,

          \text { slope of normal }=\frac{-1}{\text { slope of tangent }}

Given :

Here given the curve,

            x=\frac{1}{t} \text { and } y=t

We have to find the slope of normal.

Solution :

Here,

           x=\frac{1}{t} \text { and } y=t

Differentiating with respect to 't', we get

\begin{gathered} \Rightarrow \quad \frac{d x}{d t}=\frac{-1}{t^{2}} \text { and } \frac{d y}{d t}=1 \\ \therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{1}{\left(\frac{-1}{t^{2}}\right)} \end{gathered}

\Rightarrow \quad \frac{d y}{d x}=-t^{2}

So,        Slope of tangent =\frac{dy}{dx}= -t^{2}

              \text { slope of normal }=\frac{-1}{\text { slope of tangent }}

             =\frac{-1}{-t^{2}}=\frac{1}{t^{2}}

Hence,

\text { slope of normal }=\frac{1}{t^{2}}

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