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

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(A) Given the curve equation: $(1+x^2)y = 2-x$. To find the point where the curve crosses the $X$-axis,set $y=0$: $(1+x^2)(0) = 2-x \implies 2-x = 0 \

Vedclass · Accepted Answer

$x+5y=2$

Vedclass · Answer

(A) Given the curve equation: $(1+x^2)y = 2-x$. To find the point where the curve crosses the $X$-axis,set $y=0$: $(1+x^2)(0) = 2-x \implies 2-x = 0 \implies x=2$. So,the point of contact is $(2, 0)$. Now,differentiate the equation with respect to $x$: $(1+x^2) \frac{dy}{dx} + y(2x) = -1$. Substitute $x=2$ and $y=0$ to find the slope $m$: $(1+2^2) \frac{dy}{dx} + 0(2 	imes 2) = -1 (5) \frac{dy}{dx} = -1 \frac{dy}{dx} = -\frac{1}{5}$. The equation of the tangent at $(2, 0)$ with slope $m = -\frac{1}{5}$ is: $y - 0 = -\frac{1}{5}(x - 2) 5y = -x + 2 x + 5y = 2$. Thus,the correct option is $A$.

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

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