Question 913268
The shortest distance would be a perpendicular line from the line, {{{3x+2y+1=0}}}
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{{{2y=-3x-1}}}
{{{y=-(3/2)x-1/2}}}
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A point on the hyperbola that had the same slope would be at the minimum distance.
Find the slope of the hyperbola by differentiating,
{{{x^2/24-y^2/18=1}}}
{{{(1/24)(2xdx)-(1/18)(2ydy)=0}}}
{{{(xdx)/12=(ydy)/9}}}
{{{dy/dx=(3x)/(4y)=-3/2}}}
{{{6x=-12y}}}
{{{x=-2y}}}
Substitute into the hyperbola equation,
{{{(-2y)^2/24-y^2/18=1}}}
{{{y^2/6-y^2/18=1}}}
{{{(3/18)y^2-(1/18)y^2=1}}}
{{{(2/18)y^2=1}}}
{{{y^2=9}}}
{{{y=-3}}} and {{{y=3}}}
Then, 
{{{x^2/24-9/18=1}}}
{{{x^2/24=3/2}}}
{{{x^2=36}}}
{{{x=6}}} and {{{x=-6}}}
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{{{drawing(300,300,-10,10,-10,10,grid(1),circle(-6,3,0.3),circle(6,-3,0.3),graph(300,300,-10,10,-10,10,-(3/2)x-1/2,-sqrt(18(x^2/24-1)),sqrt(18(x^2/24-1))))}}}
So the two points are (-6,3) and (6,-3).
Now the perpendicular line that would be the shortest distance to the original line has a slope that is the negative reciprocal of the original line.
{{{m=2/3}}}
You have the slope and the point, find the line.
{{{y-3=(2/3)(x-(-6))}}}
{{{y-3=(2/3)(x+6)}}}
{{{y-3=(2/3)x+4}}}
{{{y=(2/3)x+7}}}
{{{drawing(300,300,-10,2,-2,10,grid(1),circle(-6,3,0.3),circle(6,-3,0.3),graph(300,300,-10,2,-2,10,-(3/2)x-1/2,-sqrt(18(x^2/24-1)),sqrt(18(x^2/24-1)),(2/3)x+7))}}}
Now find the intersection point of the two lines to find the other point to use to calculate minimum distance.
{{{y=-(3/2)x-1/2}}}
{{{y=(2/3)x+7}}}
{{{-(3/2)x-1/2=(2/3)x+7}}}
Multiply both side by {{{6}}}.
{{{-9x-3=4x+42}}}
{{{-13x=45}}}
{{{highlight(x=-45/13)}}}
Then,
{{{y=(2/3)(-45/13)+7}}}
{{{y=-(30/13)+91/13}}}
{{{highlight(y=61/13)}}}
So now use the distance formula,
{{{D^2=(-6-(-45/13))^2+(3-61/13)^2}}}
{{{D^2=(-78/13+45/13)^2+(39/13-61/13)^2}}}
{{{D^2=(-33/13)^2+(22/13)^2}}}
{{{D^2=(1089/169)+(484/169)}}}
{{{D^2=1573/169}}}
{{{D=sqrt(1573)/13}}}
{{{highlight(D=(11/13)sqrt(13))}}}
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You can also calculate the other point that intersects with (6,-3) and then calculate the distance.