Question 983166
A line perpendicular to {{{x-y+4=0}}} would have a slope that is the negative reciprocal of its slope.
{{{y=x+4}}}
{{{m=1}}}
{{{m[p]=-1}}}
So the slope of the perpendicular line would be {{{m=-1}}}.
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The tangent line to the hyperbola can be found by differentiating,
{{{2x-8y=0}}}
{{{2xdx=8ydy}}}
{{{m=dy/dx=(2x)/(8y)=x/(4y)}}}
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This slope must equal the previously calculated slope,
{{{x/(4y)=-1}}}
{{{4y=-x}}}
{{{y=-x/4]}}
Substitute back into the hyperbola to find the points of intersection.
{{{x^2-4(-x/4)^2=36}}}
{{{x^2(1-1/4)=36}}}
{{{(3/4)x^2=36}}}
{{{x^2=48}}}
{{{x=0 +- 4sqrt(3)}}}
So then,
{{{y=0 +- sqrt(3)}}}
The two points of intersection are,
({{{-4sqrt(3)}}},{{{sqrt(3)}}}) and ({{{4sqrt(3)}}},{{{-sqrt(3)}}}) 
So then using the point-slope form of a line,
{{{y-sqrt(3)=-(x+4sqrt(3))}}} and {{{y+sqrt(3)=-(x-4sqrt(3))}}}
{{{highlight_green(y=-x+3sqrt(3))}}} and {{{highlight_green(y=-x-3sqrt(3))}}}
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*[illustration fd7.JPG].