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Need solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Very short Answers Question 12 textbook solution.

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Answer : Equation of tangent, x+y=2

Hint: Use the equation of tangent,

         y-y_{1}=\frac{d y}{d x}\left(x-x_{1}\right)

Given : Here the curve,

             y=x^{2}-x+2  at the point where it crosses the y-axis.

We have to write the equation of the tangent.

Solution :

We know that,

When the curve passes through  y - axis, then the point on the curve is of the form (0,y)

Now,   y=x^{2}-x+2

Differentiating with respect to x,

             \frac{d y}{d x}=2 x-1

           Slope of tangent =\left(\frac{d y}{d x}\right)_{(0,2)}=2(0)-1

                                     =-1= m (say)

\therefore \quad\left(x_{1}, y_{1}\right)=(0,2)_{\text {and }} m=1

Equation of tangent,

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

Hence the required equation of tangent, x+y = 2.




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