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What is the smallest distance between the origin and a point on the graph of y = 1/2*(x^2 - 8)?

 Nov 19, 2022
 #1
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The smallest possible distance is 5*sqrt(5).

 Nov 19, 2022
 #2
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y = 1/2x^2 - 4

 

Suppose that \((x_0, y_0)\) is a point on this graph closest to the origin.

 

Hence:

\(y_0 = \frac12(x_0)^2 - 4.\)

 

Using this information, we need to find 

\(\sqrt{x_0^2 + y_0^2}\)

 

We can plug in \(x_0^2\) with \(2y_0 + 8\) (Make sure you see how I got this)

 

Hence, the distance between the points will be:

\(\sqrt(y_0^2 +2y_0 + 8)\)

Since this distance is to be minimized by how we defined the point (x_0, y_0), we can find the vertex of the that quadratic in y_0. If you complete the square, you will recieve that the parabola reaches its minimum when \(y_0 = -1\).

Can you solve the problem from here?

 Nov 19, 2022
 #3
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I don't understand.  Can you explain more?

Guest Nov 26, 2022

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