Question

Find the point on the hyperbola \(xy = 8\) that is closest to the point \(\left( {3,0} \right)\).

Short answer

\(\left( {4,2} \right)\) is the point on hyperbola that is closest to \(\left( {3,0} \right)\).

Step-by-step solution

Modify the given equation

For a minimum value of distance, differentiate the above equation and equate to zero gives,

\(\begin{aligned}{c}2dd' &= 0\\2x - 6 - \frac{{128}}{{{x^3}}} &= 0\\{x^4} - 3{x^3} - 64 &= 0\\x &= 4, - 2.949\end{aligned}\)

Take \(x = 4\) and neglect negative value.

Thus, the point is obtained as:

\(\begin{aligned}{c}\left( {x,\frac{8}{x}} \right) &= \left( {4,\frac{8}{4}} \right)\\ &= \left( {4,2} \right)\end{aligned}\)

Hence, \(\left( {4,2} \right)\) is the point on hyperbola that is closest to \(\left( {3,0} \right)\).

Calculate the distance

Let the distance of \(\left( {x,\frac{8}{x}} \right)\) from the given point \(\left( {3,0} \right)\) on the hyperbola.

The distance is calculated as follows:

\(\begin{aligned}{c}d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \\ = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {\frac{8}{x} - 0} \right)}^2}} \\ = \sqrt {{x^2} - 6x + 9 + \frac{{64}}{{{x^2}}}} \end{aligned}\)

Solve further the above equation gives,

\(\begin{aligned}{l}d = \sqrt {{x^2} - 6x + 9 + \frac{{64}}{{{x^2}}}} \\{d^2} = {x^2} - 6x + 9 + \frac{{64}}{{{x^2}}}\end{aligned}\)

Obtain the minimum distance value of the hyperbola

For a minimum value of distance, differentiate the above equation and equate to zero gives,

\(\begin{aligned}{c}2dd' &= 0\\2x - 6 - \frac{{128}}{{{x^3}}} &= 0\\{x^4} - 3{x^3} - 64 &= 0\\x &= 4, - 2.949\end{aligned}\)

Take \(x = 4\) and neglect negative value.

Thus, the point is obtained as:

\(\begin{aligned}{c}\left( {x,\frac{8}{x}} \right) &= \left( {4,\frac{8}{4}} \right)\\ &= \left( {4,2} \right)\end{aligned}\)

Hence, \(\left( {4,2} \right)\) is the point on hyperbola that is closest to \(\left( {3,0} \right)\).