Find the equation of the normal to the curve | vedclass

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(D) Given curve is $2y = 3 - x^2$.Differentiating with respect to $x$,we get $2 \frac{dy}{dx} = -2x$,which implies $\frac{dy}{dx} = -x$.The slope of t

Vedclass · Accepted Answer

$x - y = 0$

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(D) Given curve is $2y = 3 - x^2$.Differentiating with respect to $x$,we get $2 \frac{dy}{dx} = -2x$,which implies $\frac{dy}{dx} = -x$.The slope of the tangent at $(1, 1)$ is $m = \left(\frac{dy}{dx}\right)_{(1,1)} = -1$.The slope of the normal at $(1, 1)$ is $-\frac{1}{m} = -\frac{1}{-1} = 1$.The equation of the normal at $(1, 1)$ is $(y - 1) = 1(x - 1)$.Simplifying this,we get $y - 1 = x - 1$,which results in $x - y = 0$.

Find the equation of the normal to the curve $2y = 3 - x^2$ at the point $(1, 1)$.

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