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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equations of the tangent and normal to the given curves at the indicated points: $ (ii) \: y = x^4 - 6x^3 + 13x^2 - 10x + 5\; at \;(1, 3)$

This is second part of the multi-part question q14

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1 Answer

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Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Given :$y=x^4-6x^3+13x^2-10x+5$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=$$4x^3-18x^2+26x-10$
The given points are $(1,3)$
Put $x=1$
$\large\frac{dy}{dx}$$=4(1)-18(1)+26(1)-10$
$\quad\;=4-18+26-10$
$\large\frac{dy}{dx}$$=2$
Therefore slope of the tangent at $(1,3)$ is $2$
Step 2:
Therefore equation of the tangent at $(1,3)$ is $(y-3)=2(x-1)$
$\Rightarrow y-3=2x-2$
$\Rightarrow 2x-y+1=0$
Step 3:
Equation of the normal at $(1,3)$ is $(y-3)=\large\frac{-1}{2}$$(x-1)$
$\Rightarrow 2y-6=-x+1$
$\Rightarrow x+2y-7=0$
answered Jul 11, 2013 by sreemathi.v
 

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