Question 364660
You can get the value of the slope of the tangent line for the curve by taking the derivative. 
{{{m[t]=-2/x^2}}}
The tangent and the normal to the curve are perpendicular to each other. 
PErpendicular lines have slopes that are negative reciprocals.
{{{m[t]*m[n]=-1}}}
{{{m[n]=-1/m[t]}}}
{{{m[n]=x^2/2}}}
Now you have the slope of the normal line and you have one point (0,-3).
Use the point slope form of a line,{{{y-y[p]=m(x-x[p])}}}
{{{y-(-3)=(x^2/2)(x-0)}}}
{{{y+3=x^3/2}}}
{{{y=x^3/2-3}}}
But you also know that,
{{{y=2/x}}}
Substitute,
{{{2/x=x^3/2-3}}}
Multiply both sides by {{{2x}}},
{{{4=x^4-6x}}}
{{{x^4-6x-4=0}}}
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{{{graph(300,300,-2,5,-2,3,x^4-6x-4)}}}
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That curve has a zero at {{{x=2}}}
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{{{drawing(300,300,-2,8,-5,5,circle(2,1,0.2),locate(2.2,1.8,P),circle(0,-3,0.2),blue(line(2,1,0,-3)),grid(1),graph(300,300,-2,8,-5,5,-(1/2)(x-2)+1,2/x))}}}
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P:(2,1)