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  • Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 5 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 5 Maths Textbook Solution.

Answers (1)

Answer: Equation of tangent, y+2 x=2

              Equation  of normal , 2 y-x+6=0

HINTS:

 Differentiating the given equation with respect to x.

Given:

y^{2}=\frac{x^{3}}{4-x} \text { at }(2,-2)

Solution:

2 y \frac{d y}{d x}=\frac{(4-x) 3 x^{2}+x^{3}}{(4-x)^{2}}

    \frac{d y}{d x}=\frac{(4-x) 3 x^{2}+x^{3}}{2 y(4-x)^{2}}

m(\text { tangent }) \text { at }(2,-2)=-2

     \mathrm{m} \text { (normal) at }(3,2)=-\frac{1}{2}

Equation of tangent is,

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

\therefore y+2 x-2=0

Equation of Normal  is,

y_{1}=m(\text { normal })\left(x-x_{1}\right)

  y+2=\left(-\frac{1}{2}\right)(x-2)

2 y+4=-x+2

 \therefore 2 y+x+2=0

 

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