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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