Find the minimum distance from the point (0, | vedclass

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(C) Let $P(x, y)$ be a point on the parabola $y = x^2$ and $Q(0, c)$ be the given point.Since $P$ lies on the parabola,$y = x^2$. The distance $PQ$ is

Vedclass · Accepted Answer

$\frac{\sqrt{4c - 1}}{2}$

Vedclass · Answer

(C) Let $P(x, y)$ be a point on the parabola $y = x^2$ and $Q(0, c)$ be the given point.Since $P$ lies on the parabola,$y = x^2$. The distance $PQ$ is given by $PQ = \sqrt{(x - 0)^2 + (y - c)^2}$.To minimize the distance,we minimize $Z = PQ^2 = x^2 + (x^2 - c)^2$.$Z = x^2 + x^4 - 2cx^2 + c^2 = x^4 + x^2(1 - 2c) + c^2$.Taking the derivative with respect to $x$: $\frac{dZ}{dx} = 4x^3 + 2x(1 - 2c) = 2x(2x^2 + 1 - 2c)$.Setting $\frac{dZ}{dx} = 0$,we get $x = 0$ or $x^2 = \frac{2c - 1}{2}$.If $c \leq \frac{1}{2}$,then $x = 0$ is the only real solution,and the minimum distance is at $x = 0$,$PQ = \sqrt{0^2 + (0 - c)^2} = c$.If $c > \frac{1}{2}$,then $x^2 = \frac{2c - 1}{2}$ gives the minimum. Substituting $x^2 = \frac{2c - 1}{2}$ into $Z$:$Z = (\frac{2c - 1}{2})^2 + (\frac{2c - 1}{2} - c)^2 = \frac{(2c - 1)^2}{4} + (\frac{2c - 1 - 2c}{2})^2 = \frac{4c^2 - 4c + 1}{4} + \frac{1}{4} = \frac{4c^2 - 4c + 2}{4} = c^2 - c + \frac{1}{2}$.Wait,re-evaluating $Z = x^2 + (x^2 - c)^2$ at $x^2 = c - \frac{1}{2}$:$Z = (c - \frac{1}{2}) + (c - \frac{1}{2} - c)^2 = c - \frac{1}{2} + \frac{1}{4} = c - \frac{1}{4} = \frac{4c - 1}{4}$.Thus,$PQ = \sqrt{\frac{4c - 1}{4}} = \frac{\sqrt{4c - 1}}{2}$.

Find the minimum distance from the point $(0, c)$ to the parabola $y = x^2$,where $0 \leq c \leq 5$.

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