Question

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

Answer 1

Find the point on the given hyperbola:

Given that,

The hyperbola equation $\mathrm{xy}=8$
$\Rightarrow \mathrm{y}=\frac{8}{\mathrm{x}}$
The objective is to find the point on the above hyperbola, which is closest to the point $(3,0)$

Step-1: Apply the distance formula.

Let $(\mathrm{x},\mathrm{y})$ be a point $\mathrm{o}$ the hyperbola, which is closest to the point $(3,0)$.

Since, for every point $(\mathrm{x},\mathrm{y})$ on the above hyperbola, $\mathrm{y}=\frac{8}{\mathrm{x}}$.
Then, $(\mathrm{x},\mathrm{y})=\left(\mathrm{x},\frac{8}{\mathrm{x}}\right)$ is the point on the hyperbola, which is closest to the point $(3,0)$.

Apply the distance formula between the two points $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ and $\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$:

$\Rightarrow \mathrm{D}=\sqrt{{\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}^2+{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2}$.

Substitute ${\mathrm{x}}_1=\mathrm{x},{\mathrm{y}}_1=\frac{8}{\mathrm{x}},{\mathrm{x}}_2=3,\mathrm{and}{\mathrm{y}}_2=0$ in the distance formula:

$\begin{array}{c}\Rightarrow \mathrm{D}=\sqrt{(\mathrm{x}-3{)}^2+{\left(\frac{8}{\mathrm{x}}-0\right)}^2}\\ \Rightarrow \mathrm{D}=\sqrt{(\mathrm{x}-3{)}^2+{\left(\frac{8}{\mathrm{x}}\right)}^2}\\ \Rightarrow \mathrm{D}=\sqrt{{\mathrm{x}}^2-6\mathrm{x}+9+{\left(\frac{8}{\mathrm{x}}\right)}^2}\left[\mathrm{apply}{\left(\mathrm{a}-\mathrm{b}\right)}^2={\mathrm{a}}^2-2\mathrm{ab}+{\mathrm{b}}^2\right]\\ \Rightarrow \mathrm{D}=\sqrt{{\mathrm{x}}^2-6\mathrm{x}+9+\frac{64}{{\mathrm{x}}^2}}\left[\mathrm{Expand}{\left(\frac{8}{\mathrm{x}}\right)}^2\right]\end{array}$

Step-2: Apply the Differentiation to find $\mathbf{x}$ value.

For extreme value of $\mathrm{D},$ the value of ${\mathrm{D}}^{\text{'}}=0$.

That is, the distance $\mathrm{D}$ is minimum, only when ${\mathrm{D}}^{\text{'}}=0$.

Since, $\mathrm{D}=\sqrt{{\mathrm{x}}^2-6\mathrm{x}+9+\frac{64}{{\mathrm{x}}^2}}$.
Differentiate the term with respect to $\mathrm{x}$:
$\Rightarrow {\mathrm{D}}^{\text{'}}=\frac{1}{2\sqrt{{\mathrm{x}}^2-6\mathrm{x}+9+\frac{64}{{\mathrm{x}}^2}}}\left(2\mathrm{x}-6+0-\frac{128}{{\mathrm{x}}^3}\right)$

Cross multiply the term ${\mathrm{x}}^2$ in the term $2\sqrt{{\mathrm{x}}^2-6\mathrm{x}+9+\frac{64}{{\mathrm{x}}^2}}$:

$\Rightarrow {\mathrm{D}}^{\text{'}}=\frac{1}{{\displaystyle \frac{2}{\mathrm{x}}}\sqrt{{\mathrm{x}}^4-6{\mathrm{x}}^3+9{\mathrm{x}}^2+64}}\left(2\mathrm{x}-6+0-\frac{128}{{\mathrm{x}}^3}\right)$

$\begin{array}{c}\Rightarrow \mathrm{D}\text{'}=\frac{1}{\frac{2}{\mathrm{x}}\sqrt{{\mathrm{x}}^4-6{\mathrm{x}}^3+9{\mathrm{x}}^2+64}}\left(\frac{2{\mathrm{x}}^4-6{\mathrm{x}}^3-128}{{\mathrm{x}}^3}\right)\end{array}$

Multiply the expression:

$\begin{array}{c}\Rightarrow \mathrm{D}\text{'}=\frac{2\left({\mathrm{x}}^4-3{\mathrm{x}}^3-64\right)}{2{\mathrm{x}}^2\sqrt{{\mathrm{x}}^4-6{\mathrm{x}}^3+9{\mathrm{x}}^2+64}}\\ \Rightarrow \mathrm{D}\text{'}=\frac{{\mathrm{x}}^4-3{\mathrm{x}}^3-64}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^4-6{\mathrm{x}}^3+9{\mathrm{x}}^2+64}}\end{array}$

Step-3: Find the $\mathbf{x}$ value.

Substitute ${\mathrm{D}}^{\text{'}}=0$:

$\Rightarrow 0=\frac{{\mathrm{x}}^4-3{\mathrm{x}}^3-64}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^4-6{\mathrm{x}}^3+9{\mathrm{x}}^2+64}}$

Rewrite the equation:
$\begin{array}{r}\Rightarrow \frac{{\mathrm{x}}^4-3{\mathrm{x}}^3-64}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^4-6{\mathrm{x}}^3+9{\mathrm{x}}^2+64}}=0\\ {\mathrm{x}}^4-3{\mathrm{x}}^3-64=0\\ (\mathrm{x}-4)\left({\mathrm{x}}^3+{\mathrm{x}}^2+4\mathrm{x}+16\right)=0\\ \mathrm{x}-4=0\\ \mathrm{x}=4\end{array}$
Step-4: Find the $\mathbf{y}$ value.

Substitute $\mathrm{x}=4$ in the equation $\mathrm{y}=\frac{8}{\mathrm{x}}$:

$\begin{array}{rcl}& \Rightarrow & \mathrm{y}=\frac{8}{4}\\ & \Rightarrow & \mathrm{y}=2\end{array}$

So, the values of $\left(\mathrm{x},\mathrm{y}\right)$ are $\left(4,2\right).$

Hence, the point on the hyperbola $\mathbf{xy}\mathbf{=}\mathbf{8}$ that is closest to the point $\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}$ is $\left(4,2\right).$