The distance \(l\) between \((x, y)\) and \((3, 0)\) is:
\(l=\sqrt{(0-y)^{2}+(3-x)^{2}}=\sqrt{y^{2}+(3-x)^{2}}\)
Since the equation of the hyperbola is \(xy=8\), we can get \(y\) by itself and end up with
\(y=\frac{8}{x}\)
which we can plug into our distance formula:
\(l=\sqrt{(\frac{8}{x})^{2}+(3-x)^{2}}\)
To make calculation easier, we'll square both sides:
\(l^{2}=(\frac{8}{x})^2+(3-x)^{2}\)
and create a new variable \(m=l^{2}\):
\(m=(\frac{8}{x})^2+(3-x)^{2}\)
\(\rightarrow m=\frac{64}{x^{2}}+9-6x+x^{2}\)
Differentiate both sides:
\(\frac{dm}{dx}=-\frac{128}{x^{3}}-6+2x\)
Minimum distance is achieved when \(\frac{dm}{dx}=0\):
\(-\frac{128}{x^{3}}-6+2x=0\)
\(\rightarrow -128-6x^{3}+2x^{4}=0\)
\(\rightarrow 2(x^{4}-3x^{3}-64)=0\)
\(\rightarrow x^{4}-3x^{3}-64=0\)
To find a value of \(x\), you can use methods like synthetic division and get the answer \(x=4\)
Plug into \(xy=8\):
\(4y=8\)
\(\rightarrow y=2\)
So the closest point on the hyperbola to \((3, 0)\) is \((4, 2)\)
