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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 16 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 16 Maths Textbook Solution.

Answers (1)

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

                 Equation  of normal , y+x=3

HINTS:

 Differentiating the given curve with respect to x .

GIVEN:

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

SOLUTION:

\begin{aligned} 2 y \frac{d y}{d x} &=4 \\ \frac{d y}{d x_{(1,2)}} &=\frac{4}{2 y} \\ &=\frac{2}{y}=1 \end{aligned}

Slope of tangent =\frac{dy}{dx}=1

Slope of normal = -\frac{1}{\text { slope of tangent }}=-1

Equation of tangent is

\begin{aligned} &y-2=1(x-1) \\ &\therefore x-y+1=0 \end{aligned}

Equation of Normal  is ,

\begin{aligned} &y-2=-1(x-1) \\ &\therefore y+x=3 \end{aligned}

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