Find the point on the parabola y = x 2 nearest to the point (-3,0)?

Question

Find the point on the parabola   y  =      x    2   nearest to the point (-3,0)?

usifuna7qo · Accepted Answer

The distance from any point (x,y) to a point

(\hat{x}, \hat{y})

is

XXX

\sqrt{{(x - \hat{x})}^{2} + {(y - \hat{y})}^{2}}

For points on

y = x^{2}

this becomes

XXX d (x) = \sqrt{{(x - \hat{x})}^{2} + {(x^{2} - \hat{y})}^{2}}

and more specifically for the point

(\hat{x}, \hat{y}) = (- 3, 0)

this becomes

XXX \sqrt{{(x + 3)}^{2} + {(x^{2} - 0)}^{2}}

XX = \sqrt{x^{4} + x^{2} + 6 x + 9}

The issue is to reduce d(x), or alternatively, and more simply, to reduce

XXX f (x) = x^{4} + x^{2} + 6 x + 9

The minimum occurs when

f^{'} (x) = 0

That is when

XXX 4 x^{3} + 2 x + 6 = 0

An obvious (by inspection) root is x=-1
(and in fact there are no other real roots)
If x=-1
then

y = x^{2} = {(- 1)}^{2} = 1

gibbokf5 · Accepted Answer

The squared distance from (x,x2) to (-3,0) isq=(x−−3)2+(x2−0)2=x4+x2+6x+9We want the minimum q,0=dqdx=4x3+2x+6It&#039;s a cubic, but it&#039;s a simple one. Using the tried-and-true method of testing small numbers, we discover x=−1 is a solution so x+1 is a factor.0=(x+1)(4x2−4x+6)The quadratic has no real roots, so (-1,1) is the sole solution.

Find the point on the parabola y=x^2 nearest to the point (-3,0)?

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