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What is the smallest distance between the origin and a point on the graph of $$y=\dfrac{1}{\sqrt{2}}\left(x^2-8\right)$$?

May 20, 2022

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Let $$(x, y) = \left(x, \dfrac1{\sqrt2}(x^2 - 8)\right)$$ be a point on the graph.

Then the distance between the origin to that point is $$\sqrt{x^2 + \left(\dfrac1{\sqrt 2}(x^2- 8)\right)^2} = \sqrt{x^2 + \dfrac{(x^2 - 8)^2}2} = \sqrt{\dfrac{x^4 - 14x^2 + 64}2}$$

To minimize the distance, we need to minimize $$x^4 - 14x^2 + 64$$. Note that $$x^4 - 14x^2 + 64 = (x^2 - 7)^2 + 15 \geq 15$$, with equality attained when x = sqrt(7) or x = -sqrt(7). Then the minimum distance is $$\sqrt{\dfrac{15}2} = \dfrac{\sqrt{30}}2$$.

May 20, 2022
edited by MaxWong  May 20, 2022
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Thank you. I can't believe I didn't see that before.

Guest May 20, 2022
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You're welcome.

MaxWong  May 20, 2022