Question 1055178
OK, assume that the shortest distance line segment intersects the first curve at ({{{a}}},{{{a^2+4}}}) and intersects the second curve at ({{{b^2}}},{{{b}}}).
The slope of the tangent line of the first curve at point {{{a}}} would be {{{2a}}} since {{{dy/dx=2x}}}.
The slope of the tangent line of the second curve at point {{{b}}} would be {{{1/(2b)}}} since {{{dy/dx=1/(2*sqrt(x))}}}.
The slopes are equivalent which leads to the first equation,
{{{2a=1/(2b)}}}
{{{4ab=1}}}
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Now use the distance formula (actually use the distance squared to eliminate the square root) with the two points of intersection,
{{{D^2=(a-b^2)^2+(a^2+4-b)^2}}}
From the previous equation,
{{{a=1/(4b)}}}
Substitute,
{{{D^2=(1/(4b)-b^2)^2+(1/(16b)+4-b)^2}}}
Graphing and finding the minimum,
{{{b=1.188}}}
So then,
{{{a=1/(4b)}}}
{{{a=0.210438}}}
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*[illustration lp2.JPG].
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So then the points are ({{{0.210}}},{{{4.044}}}) and ({{{1.188}}},{{{1.411}}})
Using the distance formula,
{{{D[min]=3.098}}}
You could also have just used the graphed value when we minimized the distance squared,
{{{D[min]^2=9.601}}}
{{{D[min]=3.098}}}
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*[illustration lp3.JPG].