What is the equation of the normal to the cur | vedclass

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Question

(A) Given the curve $y = x(2 - x) = 2x - x^2$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2 - 2x$.The slope of the tangent at $(2, 0)$

Vedclass · Accepted Answer

$x - 2y = 2$

Vedclass · Answer

(A) Given the curve $y = x(2 - x) = 2x - x^2$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2 - 2x$.The slope of the tangent at $(2, 0)$ is $m_t = \left( \frac{dy}{dx} \right)_{(2, 0)} = 2 - 2(2) = 2 - 4 = -2$.The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{-2} = \frac{1}{2}$.The equation of the normal at $(2, 0)$ is $y - y_1 = m_n(x - x_1)$.Substituting the values,$y - 0 = \frac{1}{2}(x - 2)$.$2y = x - 2$,which simplifies to $x - 2y = 2$.

What is the equation of the normal to the curve $y = x(2 - x)$ at the point $(2, 0)$?

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