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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The equation of normal to the curve y=tan x at (0,0) is___________.

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Toolbox:
  • Slope of a line is $ ax+by+c=0$ is $ - \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
  • If $ y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p = $ slope of the normal to $ y = f(x)$ at point $p$
Step 1
$ y = \tan \: x$
differentiating w.r.t $x$ we get,
$ \large\frac{dy}{dx}=\sec^2x$
The slope of the tangent at $(0,0)$ is
$ \large\frac{dy}{dx_{(0,0)}}= \sec^2 (0)$
But $ \sec 0=1$
$ \therefore$ slope of the tangent is m=1$
Slope of the normal is $ -\large\frac{1}{m}=-1$
Step 2
Equation of the normal is
$ y-y_1= -\large\frac{1}{m} (x-x_1)$
$ \Rightarrow (y-0)=-1(x-0)$
$ x+y=0$
Equation of the normal is $ x+y=0$
answered Aug 13, 2013 by thanvigandhi_1
 
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