Chapter 13.6, Problem 18PSC

To determine

To find: The equation of the point such that its distance from the point $\left(3,0\right)$ is twice its distance from the point $\left(-3,0\right)$ .

Answer to Problem 18PSC

The required equation is ${\left(x+5\right)}^2+y^2=16$ .

Explanation of Solution

Calculation:

Consider the required diagram is shown in Figure 1

  

Figure 1

Consider the distance between the point A and P is,

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

The distance between the point B and P is,

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

The distance between BP is twice the distance between AP as,

  $\begin{array}{l}2\sqrt{{\left(x+3\right)}^2+y^2}=\sqrt{{\left(x-3\right)}^2+y^2}\\ 4\left[{\left(x+3\right)}^2+y^2\right]=\left[{\left(x-3\right)}^2+y^2\right]\\ 4x^2+24x+36+4y^2=x^2-6x+9+y^2\\ x^2+10x+y^2=-9\end{array}$

Solve further as,

  $\begin{array}{l}{x}^2+10x+y^2=-9\\ x^2+10x+25+y^2=-9+25\\ {\left(x+5\right)}^2+y^2=16\end{array}$