# Find the point on the hyperbola xy = 8 that

Find the point on the hyperbola xy = 8 that is closest to the point (3,0)
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Louis Smith
Given $xy=8$
so,
$y=\frac{8}{x}$
now any point of given hyperbola can be written as $\left(x,\frac{8}{x}\right)$
so,
Let's that point on hyperbola is $\left(x,\frac{8}{x}\right)$
Now we have to find distance between $\left(x,\frac{8}{x}\right)$ and $\left(3,0\right)$, and minimize the distance
distance between two given points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ are given by
$d=\sqrt{{\left({x}^{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
so,
$d=\sqrt{{\left(x-3\right)}^{2}+{\left(\frac{8}{x}-0\right)}^{2}}$
$d=\sqrt{{\left(x-3\right)}^{2}-\frac{64}{{x}^{2}}}$
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
${d}^{2}={\left(x-3\right)}^{2}+\frac{64}{{x}^{2}}$
$2d{d}^{\prime }=2\left(x-3\right)-\frac{64}{{x}^{3}}=0$
$\frac{2\left(x-3\right){X}^{3}-64}{{x}^{3}}=0$
$=\frac{2\left(x-3\right){x}^{3}-64}{{x}^{3}}=0$
$=\frac{2\left[{x}^{3}\left(x-3\right)-32\right\}\left\{{x}^{3}\right\}=0}{}$
$={x}^{4}-3{x}^{3}-32=0$
at $x=4,{x}^{4}-3{x}^{3}-32=0$ becomes zero
so, that point will be $\left(4,\frac{8}{4}\right)\equiv \left(4,2\right)$
hence, point on hyperbola whose distance from (3,0) will be minimum is (4,2).