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The smallest distance between the origin and a point on the graph of $y=\frac{1}{2}x^2-9$ can be expressed as $a$. Find $a^2$.

I searched up how to calculate the distance between a parabola and the origin. Many of the results talked about "Lagrange Multipliers". However, I do not understand this concept. Does anyone know any other way to solve?

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  • $\begingroup$ Just minimize $$x^2+(\frac{1}{2}x^2-9)^2$$ which is the square of the distance of the point $(x/f(x))$ from the origin. $\endgroup$ – Peter Nov 23 '17 at 20:30
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    $\begingroup$ Lagrange multipliers are a technique taught in multivariable calculus, usually in the second or third semester of a college degree. It's a very powerful tool, but it's over kill for this question. $\endgroup$ – Duncan Ramage Nov 23 '17 at 20:31
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Here is the graph pf the functionenter image description here It is apparent that the minimum distance from the origin occurs about $x=\pm4$.

Here is the graph of the distance of the points on the parabola as a function of $x$

enter image description here

It is even more convincing that the minimum distance occurs very close to $\pm4$.

Let's calculate the distance as a function of $x$. At an $x$ the coordinates of a point on the graph are $(x,\frac12x^2-9)$. That is, the distance form the origin is

$$\sqrt{x^2+\left(\frac12x^2-9\right)^2}.$$ If we are after the minimum, we do do not have to take the square root. So, we want to minimize $$x^2+\left(\frac12x^2-9\right)^2=\frac14x^4-8x^2+81$$ whose derivative is

$$x^3-16x^2=x(x^2-16).$$

The solutions of the equation $x^3-16x^2=x(x^2-16)=0$ are $0,\pm 4$ as we suspected.

The minimum distance is then $\sqrt{17}.$

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The distance between two points $(x,y)$ and $(p,q)$ is defined by $\|(x,y)-(p,q)\|=\sqrt{(x-p)^2+(y-q)^2}$. Now you like to consider points from the graph where $y=\frac12x^2-9$ and $(p,q)=(0,0)$ the orgin, so you like to compute the $x$ such that $$ d(x)=\left\|\left(x,\frac12x^2-9\right)-(0,0)\right\|=\sqrt{x^2+\left(\frac12x^2-9\right)^2} $$ is minimal. To simplify the function, you should consider $$ f(x)=d^2(x)=x^2+\left(\frac12x^2-9\right)^2=\frac14x^4-8x^2+81. $$ The minimum of $f$ is then the square of the minimal distance.

This way you don't need the Lagrange multiplier. Just compute the minimum of $f$.

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The equation of the tangent line to the parabola at $(x_0,y_0)$ is: $$y_1=\frac12x_0-9+x_0(x-x_0) \Rightarrow y_1=x_0x-\frac12x_0^2-9.$$ The line passing through the origin and perpendicular to the tangent line is: $$y_2=-\frac{1}{x_0}x$$ The intersection point of the lines $y_1$ and $y_2$ is $(x_0,-1)$. Hence, from $y_1$ we find $x_0$: $$x_0^2-\frac12x_0^2-9=-1 \Rightarrow x_0=\pm4.$$ The distance between the origin and $(x_0,y_0)$ is: $$d=\sqrt{(4-0)^2+(-1-0)^2}=\sqrt{17}.$$

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