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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 3 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 3 Maths Textbook Solution.

Answers (1)

Answer:

            \left ( a \right )x-2y=2

Hint:

        Use slope of tangent m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Given:

            y=x\left ( 2-x \right )

Solution:

The given curve is y=2x-x^{2}

i.e. x^{2}-2x+y=0

Now the equation of the tangent at the point \left ( x_{1},x_{2} \right )=\left ( 2,0 \right )

Or

\begin{aligned} &x x_{1}-\left(x+x_{1}\right)+\frac{1}{2}\left(y+y_{1}\right)=0 \\ &2 x+(-1)(x+2)+\frac{1}{2}(y+0)=0 \\ &2 x-4+y=0 \\ &y=-2 x+4 \end{aligned}

So, the slope of the tangent is \left ( -2 \right )

Therefore the slope of the normal=\frac{1}{2}

So, the equation of the normal at the point \left ( 2,0 \right )is

y-0=\frac{1}{2}\left ( x-2 \right )

2y=x-2

I.e.  x-2y=2 is required

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