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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.

 Jun 23, 2023
edited by HumenBeing  Jun 23, 2023
 #1
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The point on the graph is of the form (x,21​x2−9), so the distance between the origin and this point is x2+(21​x2−9)2​. Squaring both sides, we get \begin{align*} a^2 &= x^2 + \left(\frac{1}{2}x^2-9\right)^2 \ &= x^2 + \frac{1}{4}x^4 - 9x^2 + 81 \ &= \frac{1}{4}x^4 - 8x^2 + 81. \end{align*}We can factor this as \begin{align*} a^2 &= \frac{1}{4}(x^2 - 8)^2 + \frac{324}{4} \ &= \frac{1}{4}(x^2 - 8)^2 + 81. \end{align*}The minimum value of 41​(x2−8)2 is 0, which occurs when x2=8.

Therefore, the minimum value of a^2 is 81, so a^2=81​.

 Jun 23, 2023
 #2
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HumenBeing is trying to cheat, as usual.

 Jun 23, 2023

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