#1**+2 **

**1. What is the smallest distance between the origin and a point on the graph of \(y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right)?\)**

\(\text{Let the distance $=s$} \)

\(\begin{array}{|rcll|} \hline s^2 &=& x^2+y^2 \quad &|\quad y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right) \\ s^2 &=& x^2+\left(\dfrac{1}{\sqrt{2}}\left(x^2-3\right)\right)^2 \\ s^2 &=& x^2+\dfrac{1}{2}\left(x^2-3\right)^2 \\ \dfrac{d\ s^2}{dx} &=& 2x_{\text{min}}+\dfrac{2}{2} (x_{\text{min}}^2-3)2x_{\text{min}} \\ &=& 2x_{\text{min}}+\dfrac{2}{2} (x_{\text{min}}^2-3)2x_{\text{min}} \\ &=& 2x_{\text{min}}(1+x_{\text{min}}^2-3) \\ &=& 2x_{\text{min}}(x_{\text{min}}^2-2) \\\\ \dfrac{d\ s^2}{dx} &=& 0 \\ 2x_{\text{min}}(x_{\text{min}}^2-2) &=& 0 \\ x_{\text{min}}^2-2 &=& 0 \\ \mathbf{ x_{\text{min}}^2 } &=& \mathbf{2} \\\\ y_{\text{min}} &=& \dfrac{1}{\sqrt{2}}\left(x_{\text{min}}^2-3\right) \\ y_{\text{min}} &=& \dfrac{1}{\sqrt{2}}\left(2-3\right) \\ \mathbf{ y_{\text{min}} } &=& -\dfrac{1}{\sqrt{2}} \\\\ s_{\text{min}} &=& \sqrt{x_{\text{min}}^2+y_{\text{min}}^2} \\ s_{\text{min}} &=& \sqrt{2+\left(-\dfrac{1}{\sqrt{2}}\right)^2} \\ s_{\text{min}} &=& \sqrt{ 2+ \dfrac{1}{2} } \\ s_{\text{min}} &=& \sqrt{ \dfrac{5}{2} } \\ \mathbf{ s_{\text{min}}} &=& \mathbf{1.58113883008\ldots} \\ \hline \end{array}\)

heureka Jul 22, 2019

#2**+1 **

Denoting the distance between the origin and the parabola by s,

\(\displaystyle s^{2}=x^{2}+y^{2}=x^{2}+\frac{1}{2}(x^{2}-3)^{2} \\ =x^{2}+\frac{1}{2}(x^{4}-6x^{2}+9) \\ = \frac{1}{2}(2x^{2}+x^{4}-6x^{2}+9) \\ =\frac{1}{2}(x^{4}-4x^{2}+9) \\ =\frac{1}{2}\{(x^{2}-2)^{2}+5\}.\)

Since \(\displaystyle (x^{2}-2) \ge0,\)

it follows that the least value of s^2 will be 5/2, in which case the least value of s will be sqrt(5/2) and that that happens when x^2 - 2 = 0,

ie. when x = sqrt(2).

Guest Jul 22, 2019