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Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 8 maths Textbook Solution.

Answers (1)

Answer:

            \left ( d \right )\left ( 6,36 \right )

Hint:

          Use differentiation and slope of tangent is zero

Given:

            The curve y=12x-x^{2}

Solution:

Let the point be p\left ( x,y \right )

\begin{aligned} &y=12 x-x^{2} \\ &\frac{d y}{d x}=12-2 x \\ &\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=12-2 x_{1} \end{aligned}

Since slope of tangent is zero

\begin{aligned} &\text { So, }\left(\frac{d y}{d x}\right)_{x_{1}, y_{1}}=0\\ &12-2 x_{1}=0\\ &2 x_{1}=12\\ &x_{1}=6 \end{aligned}

Also curve passing through tangent

\begin{aligned} &y_{1}=12 x_{1}-x_{1}^{2} \\ &y_{1}=12 \times 6-36 \\ &y_{1}=72-36=36 \end{aligned}

The Points are \left ( 6,36 \right )

   

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