Question

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

Answer 1

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=(8)/(x)`

which we can plug into our distance formula:

`l=\sqrt{((8)/(x))^(2)+(3-x)^(2)}`

To make calculation easier, we'll square both sides:

`l^(2)=((8)/(x))^2+(3-x)^(2)`
and create a new variable
`m=l^(2)`:

`m=((8)/(x))^2+(3-x)^(2)`

`\rightarrow m=(64)/(x^(2))+9-6x+x^(2)`

Differentiate both sides:

`(dm)/(dx)=-(128)/(x^(3))-6+2x`
Minimum distance is achieved when
`(dm)/(dx)=0`:

`-(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)`