The coordinates of the point on the parabola y^2 = 8x which is at a minimum distance from the circle x^2 + (y+6)^2 = 1 are what?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the minimum distance between the center of the circle and any point located on parabola.

Notice that the given equation of the circle indicates the coordinates of the center of the circle as `(0,-6)` . The points located on parabola have the coordinates `(x,+-sqrt(8x))` .

You need to evaluate the distance between the center of the circle and the point located on parabola, such that:

`d = sqrt((0 - x)^2 + (-6 - sqrt(8x))^2)`

`d = sqrt(x^2 + 36 + 12sqrt(8x) + 8x)`

You need to minimize the distance between the points, hence, you need to differentiate the distance function with respect to x, such that:

`d'(x) = ((x^2 + 36 + 12sqrt(8x) + 8x)')/(2(sqrt(x^2 + 36 + 12sqrt(8x) + 8x)))`

`d'(x) = (2x + (12*8)/(2sqrt(8x)) + 8)/(2(sqrt(x^2 + 36 + 12sqrt(8x) + 8x)))`

`d'(x) = (2x + 48/(sqrt(8x)) + 8)/(2(sqrt(x^2 + 36 + 12sqrt(8x) + 8x)))`

Factoring out 2 yields:

`d'(x) = (x + 24/(sqrt(8x)) + 4)/(sqrt(x^2 + 36 + 12sqrt(8x) + 8x))`

You need to solve for x the equation `d'(x) = 0` such that:

`(x + 24/(sqrt(8x)) + 4) = 0`

`(sqrt(x^2 + 36 + 12sqrt(8x) + 8x)) != 0`

Bringing the terms of equation `(x + 24/(sqrt(8x)) + 4) = 0` to a common denominator, yields:

`x(sqrt(8x)) + 24 + 4(sqrt(8x)) = 0`

You need to use the following substitution, such that:

`sqrt(8x) = t => x = t^2/8`

Changing the variable inside the equation `x(sqrt(8x)) + 24 +` `4(sqrt(8x)) = 0` yields:

`t^3/8 + 24 + 4t = 0 => t^3 + 192 + 32t = 0`

`t^3 + 4t^2 - 4t^2 + 48t - 16t + 192 = 0`

You need to group the terms such that:

`(t^3 + 4t^2) - (4t^2 + 16t) + (48t + 192) = 0`

Factoring out `t^2` in the first group, `4t` in the second group, 48 in the third group, yields:

`t^2(t + 4) - 4t(t + 4) + 48(t + 4) = 0`

Factoring out (`t + 4)` yields:

`(t + 4)(t^2 - 4t + 48) = 0`

You need to solve for t the following equations, such that:

`t + 4 = 0 => t = -4`

`t^2 - 4t + 48 = 0 => t_(1,2) = (4+-sqrt(16 - 192))/2 !in R`

You need to solve for x the equation `sqrt(8x) = -4 => 8x = (-4)^2 => x = 16/8 => x = 2`

Hence, evaluating the value of x, for the distance between the center of the circle and any point on parabola is minimized, yields `x = 2` .

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