If 0 leq x leq 5,then the minimum distance fr | vedclass

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(B) Let a point on the parabola be $(x, x^2)$. The square of the distance $d$ from $(0, c)$ is given by $z = d^2 = x^2 + (x^2 - c)^2$.Let $t = x^2$. S

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$\sqrt{c - 1/4}$

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(B) Let a point on the parabola be $(x, x^2)$. The square of the distance $d$ from $(0, c)$ is given by $z = d^2 = x^2 + (x^2 - c)^2$.Let $t = x^2$. Since $0 \leq x \leq 5$,we have $0 \leq t \leq 25$. Then $z = t + (t - c)^2 = t^2 + t(1 - 2c) + c^2$.To find the minimum,differentiate with respect to $t$: $\frac{dz}{dt} = 2t + 1 - 2c$.Setting $\frac{dz}{dt} = 0$ gives $t = c - 1/2$.If $0 \leq c - 1/2 \leq 25$ (i.e.,$1/2 \leq c \leq 51/2$),the minimum occurs at $t = c - 1/2$.The minimum distance is $\sqrt{t + (t - c)^2} = \sqrt{(c - 1/2) + (-1/2)^2} = \sqrt{c - 1/2 + 1/4} = \sqrt{c - 1/4}$.

If $0 \leq x \leq 5$,then the minimum distance from the point $(0, c)$ to the parabola $y = x^2$ is:

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