# The equation of tangent to the curve  $y(1 + x^2) = 2 - x,$ where it crosses x-axis is: $\\A. x + 5y = 2 \\B. x - 5y = 2 \\C. 5x - y = 2 \\D. 5x + y = 2$

A)

Given the equation of the curve is

$y(1 + x^2) = 2 - x,$

Both the sides are differentiated with respect to x,

$\frac{d\left(y\left(1+x^{2}\right)\right)}{d x}=\frac{d(2-x)}{d x}$
Using the power rule
$\Rightarrow y \cdot \frac{d\left(1+x^{2}\right)}{d x}+\left(1+x^{2}\right) \cdot \frac{d y}{d x}=\frac{d(2-x)}{d x}$

As the derivative of a constant is always 0 we get
$\Rightarrow y \cdot \frac{d\left(x^{2}\right)}{d x}+\left(1+x^{2}\right) \cdot \frac{d y}{d x}=\frac{d(-x)}{d x}$

Again, using the power rule
$\\\Rightarrow y \cdot 2 x+\left(1+x^{2}\right) \cdot \frac{d y}{d x}=-1 \\\Rightarrow\left(1+x^{2}\right) \cdot \frac{d y}{d x}=2 x y-1 \\\Rightarrow \frac{d y}{d x}=\frac{2 x y-1}{1+x^{2}} \ldots \ldots(i)$
The mentioned curve passes through the x -axis, i.e., y=0

Thus, the curve equation becomes

$\\ {y(1+x\textsuperscript{2})=2-x}\\ { \Rightarrow 0(1+x\textsuperscript{2})=2-x}\\ { \Rightarrow 0=2-x}\\ { \Rightarrow x=2}\\$

As the point of passing for the given curve is (2,0)

So the equation (i) at point (2,0) is,

$\\\Rightarrow\left(\frac{d y}{d x}\right)_{(2,0)}=\frac{2 x y-1}{1+x^{2}} \\\Rightarrow\left(\frac{d y}{d x}\right)_{(2,0)}=\frac{2(2)(0)-1}{1+(2)^{2}} \\\Rightarrow\left(\frac{d y}{d x}\right)_{(2,0)}=\frac{0-1}{1+4} \\\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,0)}=-\frac{1}{5}$
So, the slope of tangent to the curve is $-\frac{1}{5}$
Therefore, the equation of tangent of the curve passing through (2,0) is given by
$y-0=-\frac{1}{5}(x-2)$

$\\ { \Rightarrow 5y=-x+2}\\ { \Rightarrow x+5y=2}\\$

Thus, the equation of tangent to the curve $y(1 + x^2) = 2 - x,$, where it crosses x-axis is$x+5y=2$.

Hence, the correct option is option A