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  • Please solve RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question , Question 21 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question , Question 21 maths textbook solution. 

Answers (1)

Answer :  (b) is the correct option

Hint : \tan \theta=\left|\frac{m_{1}-1}{1-m_{1}}\right|

Given : y=6 x-x^{2}

Solution :

y=6 x-x^{2}                                              (1)

Differentiating (1) w.r.t x we get

\frac{d y}{d x}=6-2 x=m_{1}

And x+y=2                                           (2)

Since

\begin{aligned} &\tan \theta=\left|\frac{m_{1}+1}{1-m_{1}}\right| \\ &\tan \frac{\pi}{4}=\left|\frac{m_{1}+1}{1-m_{1}}\right| \\ &1=\frac{m_{1}+1}{1-m_{1}} \end{aligned}

m_{1}+1=\pm\left(1-m_{1}\right)

Taking

\begin{aligned} &m_{1}+1=1-m_{1} \\ &2 m_{1}=0 \\ &m_{1}=0 \\ &6-2 x=0 \\ &x=3 \end{aligned}

Putting the values of x in (1) we get y=9.

\therefore Point is (3,9)

 

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