Let y=f(x) be any curve on the X-Y plane and | vedclass

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(A) Let $P(x, y)$ be a point on the curve $y=f(x)$ and $C(a, b)$ be a fixed point not on the curve.The distance $d$ between $P$ and $C$ is given by $d

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$PC$ is perpendicular to the tangent at $P$

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(A) Let $P(x, y)$ be a point on the curve $y=f(x)$ and $C(a, b)$ be a fixed point not on the curve.The distance $d$ between $P$ and $C$ is given by $d^2 = (x-a)^2 + (y-b)^2$.For $d$ to be a maximum or minimum,$d^2$ must also be a maximum or minimum.Differentiating $d^2$ with respect to $x$,we get $\frac{d}{dx}(d^2) = 2(x-a) + 2(y-b)\frac{dy}{dx} = 0$.This implies $(x-a) + (y-b)\frac{dy}{dx} = 0$,which can be written as $\frac{y-b}{x-a} = -\frac{1}{dy/dx}$.Here,$\frac{y-b}{x-a}$ is the slope of the line segment $PC$,and $\frac{dy}{dx}$ is the slope of the tangent at $P$.Since the product of their slopes is $-1$,the line segment $PC$ is perpendicular to the tangent at $P$.

Let $y=f(x)$ be any curve on the $X-Y$ plane and $P$ be a point on the curve. Let $C$ be a fixed point not on the curve. If the length $PC$ is either a maximum or a minimum,then:

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