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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 15 Maths Textbook Solution.

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ANSWER: Equation of tangent, x-y-1=0

                 Equation  of normal , x+y-3=0



 Differentiating the given curve with respect to x and find its slope .


x^{2}=4 y \text { at }(2,1)


\begin{aligned} &2 x=4 \frac{d y}{d x} \\ &\Rightarrow \frac{d y}{d x}=\frac{x}{2} \end{aligned}

Slope of tangent,

m=\left(\frac{d y}{d x}\right)_{(2,1)}=\frac{2}{2}=1

Equation of tangent is

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

Equation of Normal  is ,

\begin{aligned} &y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-1=1(-x+2) \\ &\Rightarrow y-1=-x+2 \\ &\therefore x+y-3=0 \end{aligned}

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