The shortest distance between the point left( | vedclass

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Question

(A) Let $P(x, y)$ be a point on the curve $y = \sqrt{x}$. Then $P$ can be represented as $(t^2, t)$ for $t > 0$.The squared distance $D^2$ between $P(

Vedclass · Accepted Answer

$\frac{\sqrt{5}}{2}$

Vedclass · Answer

(A) Let $P(x, y)$ be a point on the curve $y = \sqrt{x}$. Then $P$ can be represented as $(t^2, t)$ for $t > 0$.The squared distance $D^2$ between $P(t^2, t)$ and $\left( \frac{3}{2}, 0 \right)$ is given by:$f(t) = (t^2 - \frac{3}{2})^2 + (t - 0)^2 = t^4 - 3t^2 + \frac{9}{4} + t^2 = t^4 - 2t^2 + \frac{9}{4}$.To find the minimum distance,we differentiate $f(t)$ with respect to $t$ and set it to zero:$f'(t) = 4t^3 - 4t = 4t(t^2 - 1) = 0$.Since $t > 0$,we have $t^2 = 1$,which gives $t = 1$.For $t = 1$,the point $P$ is $(1^2, 1) = (1, 1)$.The shortest distance is the distance between $(1, 1)$ and $\left( \frac{3}{2}, 0 \right)$:$d = \sqrt{(\frac{3}{2} - 1)^2 + (0 - 1)^2} = \sqrt{(\frac{1}{2})^2 + (-1)^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$.

The shortest distance between the point $\left( \frac{3}{2}, 0 \right)$ and the curve $y = \sqrt{x}, (x > 0)$ is:

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