The equation of the normal to the curve y(1+x | vedclass

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(A) Given curve: $y(1+x^{2})=2-x$  $(1)$At the $x$-axis,$y=0$. Substituting $y=0$ in Eq. $(1)$:$0 = 2-x \Rightarrow x=2$.So,the point of intersection

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$5x-y-10=0$

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(A) Given curve: $y(1+x^{2})=2-x$  $(1)$At the $x$-axis,$y=0$. Substituting $y=0$ in Eq. $(1)$:$0 = 2-x \Rightarrow x=2$.So,the point of intersection is $(2, 0)$.Differentiating Eq. $(1)$ with respect to $x$:$y'(1+x^{2}) + y(2x) = -1$.At point $(2, 0)$,we have:$y'(1+2^{2}) + 0(2 	imes 2) = -1$$y'(5) = -1 \Rightarrow y' = -\frac{1}{5}$.This is the slope of the tangent at $(2, 0)$.The slope of the normal is $-\frac{1}{y'} = -\frac{1}{-1/5} = 5$.The equation of the normal at $(2, 0)$ with slope $m=5$ is:$y - 0 = 5(x - 2)$$y = 5x - 10$$5x - y - 10 = 0$.

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

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