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# The equation of tangent to the curve $y(1+x^2)=2-x,$where it crosses x-axis is $\begin{array}{1 1}(A)\;x+5y=2 & (B)\;x-5y=2\\(C)\;5x-y=2 & (D)\;5x+y=2\end{array}$

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