Find the shortest distance of the point (0, c | vedclass

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Question

(A) Let $(h, k)$ be any point on the parabola $y=x^{2}$. Let $D$ be the distance between $(h, k)$ and $(0, c)$.$D = \sqrt{(h-0)^{2} + (k-c)^{2}} = \sq

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $(h, k)$ be any point on the parabola $y=x^{2}$. Let $D$ be the distance between $(h, k)$ and $(0, c)$.$D = \sqrt{(h-0)^{2} + (k-c)^{2}} = \sqrt{h^{2} + (k-c)^{2}}$.Since $(h, k)$ lies on $y=x^{2}$,we have $k=h^{2}$,so $h^{2}=k$.Substituting this into the distance formula,we get $D(k) = \sqrt{k + (k-c)^{2}}$.To find the minimum distance,we minimize $f(k) = k + (k-c)^{2} = k + k^{2} - 2kc + c^{2} = k^{2} + k(1-2c) + c^{2}$.Taking the derivative with respect to $k$,$f'(k) = 2k + 1 - 2c$.Setting $f'(k) = 0$,we get $2k = 2c - 1$,or $k = \frac{2c-1}{2}$.If $c \leq \frac{1}{2}$,the minimum distance occurs at the vertex $(0, 0)$,which is $c$. If $c > \frac{1}{2}$,the minimum distance is at $k = \frac{2c-1}{2}$.Substituting $k = \frac{2c-1}{2}$ into $D(k)$:$D = \sqrt{\frac{2c-1}{2} + (\frac{2c-1}{2} - c)^{2}} = \sqrt{\frac{2c-1}{2} + (-\frac{1}{2})^{2}} = \sqrt{\frac{2c-1}{2} + \frac{1}{4}} = \sqrt{\frac{4c-2+1}{4}} = \frac{\sqrt{4c-1}}{2}$.

Find the shortest distance of the point $(0, c)$ from the parabola $y=x^{2}$ where $0 \leq c \leq 5$.

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