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