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Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 13 Sub Question 1 Maths Textbook Solution.

Answers (1)


Equation of tangent \Rightarrow y-2x-3=0

Hint: Differentiate both with respect to x


parallel to 2x-y+9


The equation of the given curve is y=x^{2}-2 x+7

upon differentiation


This is in the form y=mx+c

\therefore slope \: \: of \: \: the\: \: line=2

If a tangent is parallel to the line 2x-y+9=0then the slope of the tangent is equal to the slope of the line. Therefore we have:

\begin{aligned} &2=2 x-2 \Rightarrow 2 x=4 \Rightarrow x=2 \\ &\text { Now, } x=2 \Rightarrow y=4-4+7=7 \end{aligned}

Thus the equation of the tangent passing through (2,7) is given by

y-7=2\left ( x-2 \right )

\Rightarrow y-2x-3=0

Hence the equation of the tangent line to the given curve (which is parallel to the line2x-y+9=0 ) is

y-2x-3=0 (Ans)

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