Graphing

What is the smallest distance between the origin and a point on the graph?

Hello Guest!

$y = \frac{1}{\sqrt{2}} (x^2 - 13)\\ y=\frac{x^2}{\sqrt{2}}-\frac{13}{\sqrt{2}}\\ y'=x\cdot\sqrt{2}\ |\ the\ perpendicular\ through\ the\ origin\ is:\\ y=-\frac{1}{\sqrt{2}}x\\ \frac{x^2}{\sqrt{2}}-\frac{13}{\sqrt{2}}=-\frac{1}{\sqrt{2}}x\ |\ \cdot \sqrt{2}\\ x^2+x-13=0$

$x\in \{-\frac{1}{2}-\frac{\sqrt{53}}{2},\color{blue}\frac{\sqrt{53}}{2}-\frac{1}{2}\}$

The smallest distance between the origin and a point on the graph

$y = \frac{1}{\sqrt{2}} (x^2 - 13)$ is 3.14005. Not correct! See answer 7#.

Here are two possible methods.

1. Algebraic.

If (x, y) is a point on he curve, then its distance from the origin d, will be such that

$d^{2}=x^{ 2}+y^{2}=x^{2}+(1/2)(x^{2}-13)^{2} \\ =x^{2}+(1/2)(x^{4}-26x^{2}+169)=(1/2)(x^{4}-24x^{2}+169) \\ =(1/2)(x^{4}-24x^{2}+144-144+169) \\ =(1/2)\{(x^{2}-12)^{2}+25\}.$

The minimum value for d^2 (and so the minimum for d), occurs when x^2 = 12.

So,

$d^{2}_{\text{min}}=25/2, \\ d_{\text{min}}=5\sqrt{2}/2 \approx 3.5355.$

2 Calculus

$d^{2}=x^{2}+y^{2}, \\ 2d\frac{d}{dx}(d)=2x+2y\frac{dy}{dx} \\ d\frac{dd}{dx} =x+\frac{1}{\sqrt{2}}(x^{2}-13)x\sqrt{2}=x(x^{2}-12).$

The minimum for d will be when dd/dx = 0, i.e. when x^2 = 12.

$x=\sqrt{12}\:\quad y=(1/\sqrt{2})(12-13)=-1/\sqrt{2}, \\ d^{2}=12+(1/2)=25/2$

as earlier, etc.

It can be seen from this graph that guest answer is most likely correct.

asinus, I am sorry but your answer is not correct.  (maybe it is just a careless error, I have not checked)