The line x + y = 2 is tangent to the curve x^ | vedclass

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(A) Given curve is $x^2 = 3 - 2y$ ...$(i)$Differentiating both sides with respect to $x$,we get:$2x = -2 \frac{dy}{dx}$$\frac{dy}{dx} = -x$The slope o

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$(1, 1)$

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(A) Given curve is $x^2 = 3 - 2y$ ...$(i)$Differentiating both sides with respect to $x$,we get:$2x = -2 \frac{dy}{dx}$$\frac{dy}{dx} = -x$The slope of the tangent to the curve at any point $(x, y)$ is $m = -x$.The given line is $x + y = 2$,which can be written as $y = -x + 2$. The slope of this line is $-1$.Since the line is tangent to the curve,the slope of the tangent must equal the slope of the line:$-x = -1$$x = 1$Substituting $x = 1$ into the equation of the curve $(i)$:$(1)^2 = 3 - 2y$$1 = 3 - 2y$$2y = 2$$y = 1$Therefore,the point of tangency is $(1, 1)$.

The line $x + y = 2$ is tangent to the curve $x^2 = 3 - 2y$ at its point

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