What is the smallest distance between the origin and a point on the graph of \( y=\dfrac{1}{\sqrt{2}}\left(x^2-8\right) \)?
Little confused. How to proceed? I was thinking find (x,y) formula for the equation, then use distance formula?
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+130071 Let the point we seek = (x , y) = (x , (x^2 -8) /sqrt 2 )
If we minimize the distance, d, then we also minimize the distance d^2
So....using the square of the distance formula between the origin and this point we have that
d^2 = ( x - 0)^2 + ( (x^2 -8) /sqrt 2 - 0 )^2 simplify
d^2 = x^2 + (1/2) ( x^4 - 16x^2 + 64 )
d^2 = x^2 + (1/2)x^4 - 8x^2 + 32
d^2 = (1/2)x^4 -7x^2 + 32 take the derivative of this function and set to 0
2x^3 - 14x = 0 factor
2x ( x^2 - 7) = 0
The second factor set to 0 is what we want
x^2 - 7 = 0
x^2 = 7
x =sqrt (7)
And y = [ (sqrt (7) )^2 - 8 ] / sqrt 2 = -1 /sqrt 2
So the smallest distance = sqrt [ (sqrt 7)^2 + (-1 /sqrt 2)^2 ] = sqrt [ 7 + 1/2] = sqrt (7.5)
Here's a graph : https://www.desmos.com/calculator/49brfcvcvv
Note that, because we have a parabola that is symmetric to the origin, (-sqrt (7) , -1/sqrt 2) also produces a point on the parabola that has the same minimum distance from the origin
