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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The equation of tangent to the curve $y(1+x^2)=2-x,$where it crosses x-axis is \[\begin{array}{1 1}(A)\;x+5y=2 & (B)\;x-5y=2\\(C)\;5x-y=2 & (D)\;5x+y=2\end{array}\]

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1 Answer

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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
The equation of the curve is
$ y(1+x^2)=2-x$
$ \Rightarrow y=\large\frac{2-x}{1+x^2}$
differentiating w.r.t $x$ we get,
$ \large\frac{dy}{dx}=\large\frac{(1+x^2)(-1)-(2-x)(1+x^2)}{(1+x^2)^2}$
$ = \large\frac{-1-x^2-(2+2x^2-x-x^3)}{(1+x^2)^2}$
$ = \large\frac{-1-x^2-2-2x^2+x+x^3}{(1+x^2)}$
$ = \large\frac{x^3-3x^2+x-3}{(1+x^2)}$
When the tangent touches the $x$ - axis
$ y=0$
$ \Rightarrow x = 2\: and \: y=0$
$ \therefore \large\frac{dy}{dx_{(2,0)}}= \large\frac{8-12+2-3}{(1+4)^2}$
$ = -\large\frac{1}{5}$
Step 2
Equation of the tangent is
$(y-0)=-\large\frac{1}{5}(x-2)$
$ 5y=-x+2$
$\Rightarrow x+5y=2$
Correct option is A
answered Aug 9, 2013 by thanvigandhi_1
 

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