The equation of the normal to the curve \( y\left(1+x^{2}\right)=2-x \) where the tangent crosses \( x \)-axis is

Given curve, $y\left(1+x^{2}\right)=2-x \rightarrow(1)$ At x-axis, $y=0$ so Eq. (1) becomes $\Rightarrow 0=2-x \Rightarrow x=2$ So, the point is $(2,0)$ Now, differentiate Eq. (1) with respect to x, we get $y^{\prime}\left(1+x^{2}\right)+y(2 x)=-1$ At point $(2,0)$, we have $\Rightarrow y^{\prime}\left(1+2^{2}\right)+0(2 \times 2)=-1$ $\Rightarrow y^{\prime}(5)=-1 \Rightarrow y^{\prime}=\frac{-1}{5}$ This is the slope of the tangent at point $(2,0)$ So, slope of normal is $-\frac{1}{y^{\prime}}=5$ Therefore, equation of normal is, $y-0=5(x-2)$ $\Rightarrow y=5 x-10$ $\Rightarrow 5 x-y-10=0$