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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: $ y = x^4 - 6x^3 + 13x^2 - 10x + 5\; at \;(0, 5)$

$\begin{array}{1 1}(A)\;10x-y+5=0;x+10y+50=0 \\ (B)\;10x+y-5=0;x-10y+50=0 \\ (C)\;-10x-y+5=0;x+10y+50=0 \\ (D)\;10x+y-5=0;-x+10y+50=0 \end{array} $

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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 $(0,5)$
Putting $x=0$
We get $\large\frac{dy}{dx}_{(0,5)}$$=-10$
Therefore slope of the tangent at $(0,5)$ is $-10$
Step 2:
Therefore equation of the tangent at $(0,5)$ is $(y-5)=-10(x-0)$
$\Rightarrow y-5=-10x$
$\Rightarrow 10x+y-5=0$
Step 3:
Therefore equation of the normal at $(0,5)$ is $(y-5)=\large\frac{-1}{10}$$(x-0)$
$\Rightarrow y-5=\large\frac{-1}{10}$$x$
$-10y+50=-x$
$\Rightarrow x-10y+50=0$
answered Jul 11, 2013 by sreemathi.v
 

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