Question

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

Answer 1

Use distance formula to get a distance function in terms of x representing the distance from a point on curve (x,y) to point (3,0)

$d = \sqrt{(x-3)^2 + y^2$

y = 8/x

$d = \sqrt{(x-3)^2 + (\frac{8}{x})^2$
Minimize the function by setting derivative equal to 0.
$\frac{d}{dx} = \frac{2(x-3) - 2(\frac{8}{x}) (\frac{8}{x^2})}{2 \sqrt{(x-3)^2 + (\frac{8}{x})^2}} = 0$
Simplify and solve for x
$x-3 = \frac{64}{x^3}$
This becomes a 4th degree polynomial, you can use graphing calculator to find zeros.
Also intuitively you can recognize that 64 = 4^3, so if x=4 you get 
1 = 1

Therefore x=4 is a solution, substituting give  y =2

Answer 2

The point that is closest to the point (3, 0) can be found by finding the

minimum value for the function of the distance between points.

The closest point on the hyperbola  x·y = 8 to the point (3, 0) is the point $\underline{(4, \ 2)}$

Reasons:

The equation of the hyperbola is x·y = 8

The given point = (3, 0)

Therefore;

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

The distance of a point from a line, $d = \mathbf{ \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}}$

Where;

(x₁, y₁) = (3, 0)

(x₂, y₂) = The closest point on the hyperbola

We get;

$d = \sqrt{\left (y-0 \right )^{2}+\left (x-3 \right )^{2}}$

Which gives;

$d = \sqrt{\left (\dfrac{8}{x} -0 \right )^{2}+\left (x-3 \right )^{2}} = \mathbf{ \sqrt{\left (\dfrac{8}{x}\right )^{2}+\left (x-3 \right )^{2}}}$

At the minimum distance, using an online app, we have;

$\dfrac{d}{dx}d = \dfrac{d}{dx} \left( \sqrt{\left (\dfrac{8}{x}\right )^{2}+\left (x-3 \right )^{2}} \right) = \dfrac{-64 + x^3 \cdot (x - 3)}{x^2 \cdot \sqrt{x^2 \cdot (x - 3)^2 + 64} } = 0$

Which gives;

x³·(x - 3) - 64 = 0

Using a graphing calculator, we get;

(x - 4)·(x³ + x² + 4·x + 16) = 0

Therefore;

A solution for the closest or point of minimum distance is x = 4

Which gives;

$y = \dfrac{8}{4} = 2$

Therefore, the closest point on the hyperbola  x·y = 8 to the point (3, 0) is the

point $\underline{(4, \ 2)}$

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