Find the point on the hyperbola xy = 8 that

Given ${\displaystyle xy=8}$
so,
${\displaystyle y=\frac{8}{x}}$
now any point of given hyperbola can be written as ${\displaystyle (x,\frac{8}{x})}$
so,
Let's that point on hyperbola is ${\displaystyle (x,\frac{8}{x})}$
Now we have to find distance between ${\displaystyle (x,\frac{8}{x})}$ and ${\displaystyle (3,0)}$, and minimize the distance
distance between two given points ${\displaystyle (x_1,y_1)}$ and ${\displaystyle (x_2,y_2)}$ are given by
${\displaystyle d=\sqrt{{(x^2-x_1)}^2+{(y_2-y_1)}^2}}$
so,
${\displaystyle d=\sqrt{{(x-3)}^2+{(\frac{8}{x}-0)}^2}}$
${\displaystyle d=\sqrt{{(x-3)}^2-\frac{64}{x^2}}}$
for finding minimum value of distance we differentiate d with respect to x and equate with zero, and find value of x where d' is zero
so, first squaring both side then differentiating
${\displaystyle d^2={(x-3)}^2+\frac{64}{x^2}}$
${\displaystyle 2d{d}^{\prime }=2(x-3)-\frac{64}{x^3}=0}$
${\displaystyle \frac{2(x-3)X^3-64}{x^3}=0}$
${\displaystyle =\frac{2(x-3)x^3-64}{x^3}=0}$
${\displaystyle =\frac{2[x^3(x-3)-32\}\left\{x^3\right\}=0}{}}$
${\displaystyle =x^4-3x^3-32=0}$
at ${\displaystyle x=4,x^4-3x^3-32=0}$ becomes zero
so, that point will be ${\displaystyle (4,\frac{8}{4})\equiv (4,2)}$
hence, point on hyperbola whose distance from (3,0) will be minimum is (4,2).