Find the point on the hyperbola xy = 8 that is closest to the point (3,0)

TokNeekCepTdh
2021-11-15
Answered

Find the point on the hyperbola xy = 8 that is closest to the point (3,0)

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Louis Smith

Answered 2021-11-16
Author has **14** answers

Given $xy=8$

so,

$y=\frac{8}{x}$

now any point of given hyperbola can be written as$(x,\frac{8}{x})$

so,

Let's that point on hyperbola is$(x,\frac{8}{x})$

Now we have to find distance between$(x,\frac{8}{x})$ and $(3,0)$ , and minimize the distance

distance between two given points$({x}_{1},{y}_{1})$ and $({x}_{2},{y}_{2})$ are given by

$d=\sqrt{{({x}^{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}$

so,

$d=\sqrt{{(x-3)}^{2}+{(\frac{8}{x}-0)}^{2}}$

$d=\sqrt{{(x-3)}^{2}-\frac{64}{{x}^{2}}}$

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

$d}^{2}={(x-3)}^{2}+\frac{64}{{x}^{2}$

$2d{d}^{\prime}=2(x-3)-\frac{64}{{x}^{3}}=0$

$\frac{2(x-3){X}^{3}-64}{{x}^{3}}=0$

$=\frac{2(x-3){x}^{3}-64}{{x}^{3}}=0$

$=\frac{2[{x}^{3}(x-3)-32\}\left\{{x}^{3}\right\}=0}{}$

$={x}^{4}-3{x}^{3}-32=0$

at$x=4,{x}^{4}-3{x}^{3}-32=0$ becomes zero

so, that point will be$(4,\frac{8}{4})\equiv (4,2)$

hence, point on hyperbola whose distance from (3,0) will be minimum is (4,2).

so,

now any point of given hyperbola can be written as

so,

Let's that point on hyperbola is

Now we have to find distance between

distance between two given points

so,

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

at

so, that point will be

hence, point on hyperbola whose distance from (3,0) will be minimum is (4,2).

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