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.
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