Solve PF_{1}=sqrt{(a-2)^2+(2/a-2)^2}

$$P F _ { 1 } = \sqrt { ( a - 2 ) ^ { 2 } + ( \frac { 2 } { a } - 2 ) ^ { 2 } }$$

Solve for F_1

$$F_{1}=-\frac{-a^{2}+2a-2}{P|a|}$$

$$P\neq 0\text{ and }a\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-2\right)^{2}.

$$PF_{1}=\sqrt{a^{2}-4a+4+\left(\frac{2}{a}-2\right)^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{a}{a}.

$$PF_{1}=\sqrt{a^{2}-4a+4+\left(\frac{2}{a}-\frac{2a}{a}\right)^{2}}$$

Since \frac{2}{a} and \frac{2a}{a} have the same denominator, subtract them by subtracting their numerators.

$$PF_{1}=\sqrt{a^{2}-4a+4+\left(\frac{2-2a}{a}\right)^{2}}$$

To raise \frac{2-2a}{a} to a power, raise both numerator and denominator to the power and then divide.

$$PF_{1}=\sqrt{a^{2}-4a+4+\frac{\left(2-2a\right)^{2}}{a^{2}}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply a^{2}-4a+4 times \frac{a^{2}}{a^{2}}.

$$PF_{1}=\sqrt{\frac{\left(a^{2}-4a+4\right)a^{2}}{a^{2}}+\frac{\left(2-2a\right)^{2}}{a^{2}}}$$

Since \frac{\left(a^{2}-4a+4\right)a^{2}}{a^{2}} and \frac{\left(2-2a\right)^{2}}{a^{2}} have the same denominator, add them by adding their numerators.

$$PF_{1}=\sqrt{\frac{\left(a^{2}-4a+4\right)a^{2}+\left(2-2a\right)^{2}}{a^{2}}}$$

Do the multiplications in \left(a^{2}-4a+4\right)a^{2}+\left(2-2a\right)^{2}.

$$PF_{1}=\sqrt{\frac{a^{4}-4a^{3}+4a^{2}+4-8a+4a^{2}}{a^{2}}}$$

Combine like terms in a^{4}-4a^{3}+4a^{2}+4-8a+4a^{2}.

$$PF_{1}=\sqrt{\frac{a^{4}-4a^{3}+4+8a^{2}-8a}{a^{2}}}$$

The equation is in standard form.

$$PF_{1}=\sqrt{\frac{a^{4}-4a^{3}+8a^{2}-8a+4}{a^{2}}}$$

Divide both sides by P.

$$\frac{PF_{1}}{P}=\frac{a^{2}-2a+2}{|a|P}$$

Dividing by P undoes the multiplication by P.

$$F_{1}=\frac{a^{2}-2a+2}{|a|P}$$

Divide \frac{2-2a+a^{2}}{|a|} by P.

$$F_{1}=\frac{a^{2}-2a+2}{P|a|}$$

Solve for P

$$P=-\frac{-a^{2}+2a-2}{F_{1}|a|}$$

$$F_{1}\neq 0\text{ and }a\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-2\right)^{2}.

$$PF_{1}=\sqrt{a^{2}-4a+4+\left(\frac{2}{a}-2\right)^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{a}{a}.

$$PF_{1}=\sqrt{a^{2}-4a+4+\left(\frac{2}{a}-\frac{2a}{a}\right)^{2}}$$

Since \frac{2}{a} and \frac{2a}{a} have the same denominator, subtract them by subtracting their numerators.

$$PF_{1}=\sqrt{a^{2}-4a+4+\left(\frac{2-2a}{a}\right)^{2}}$$

To raise \frac{2-2a}{a} to a power, raise both numerator and denominator to the power and then divide.

$$PF_{1}=\sqrt{a^{2}-4a+4+\frac{\left(2-2a\right)^{2}}{a^{2}}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply a^{2}-4a+4 times \frac{a^{2}}{a^{2}}.

$$PF_{1}=\sqrt{\frac{\left(a^{2}-4a+4\right)a^{2}}{a^{2}}+\frac{\left(2-2a\right)^{2}}{a^{2}}}$$

Since \frac{\left(a^{2}-4a+4\right)a^{2}}{a^{2}} and \frac{\left(2-2a\right)^{2}}{a^{2}} have the same denominator, add them by adding their numerators.

$$PF_{1}=\sqrt{\frac{\left(a^{2}-4a+4\right)a^{2}+\left(2-2a\right)^{2}}{a^{2}}}$$

Do the multiplications in \left(a^{2}-4a+4\right)a^{2}+\left(2-2a\right)^{2}.

$$PF_{1}=\sqrt{\frac{a^{4}-4a^{3}+4a^{2}+4-8a+4a^{2}}{a^{2}}}$$

Combine like terms in a^{4}-4a^{3}+4a^{2}+4-8a+4a^{2}.

$$PF_{1}=\sqrt{\frac{a^{4}-4a^{3}+4+8a^{2}-8a}{a^{2}}}$$

The equation is in standard form.

$$F_{1}P=\sqrt{\frac{a^{4}-4a^{3}+8a^{2}-8a+4}{a^{2}}}$$

Divide both sides by F_{1}.

$$\frac{F_{1}P}{F_{1}}=\frac{a^{2}-2a+2}{|a|F_{1}}$$

Dividing by F_{1} undoes the multiplication by F_{1}.

$$P=\frac{a^{2}-2a+2}{|a|F_{1}}$$

Divide \frac{2-2a+a^{2}}{|a|} by F_{1}.

$$P=\frac{a^{2}-2a+2}{F_{1}|a|}$$