Get Answers to all your Questions

header-bg qa

Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 1 Maths Textbook Solution.

Answers (1)

ANSWER: Equation of tangent, y+10 x-5=0

                 Equation  of normal , x-10 y+50=0


 Differentiating the given equation and find the slope of the tangent.


y=x^{4}-b x^{3}+13 x^{2}-10 x+5 \text { at }(0,5)


\frac{d y}{d x}=4 x^{3}-3 b x^{2}+26 x-10

\text { Slope of the tangent } m=\left(\frac{d y}{d x}\right)_{(0,5)}=-10

Equation of tangent is,

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y-5=-10(x-0) \\ &\Rightarrow y+10 x-5=0 \end{aligned}

Equation of Normal  is,

\begin{aligned} &y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-5=\frac{1}{10}(x-0) \\ &\Rightarrow 10 y-50=x \\ &\Rightarrow x-10 y+50=0 \end{aligned}


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support