Chapter 14.1, Problem 29PSC

To determine

To write an equation for the locus of given points. each of which is twice as far from $(-2,0)$ as it is from $(1,0)$

Answer to Problem 29PSC

  ${(x-2)}^2+y^2=4$

Explanation of Solution

Given:

The points are twice as far from $(-2,0)$ as it is from $(1,0)$

Calculation:

The following is the equation for the locus of the points-

  $\begin{array}{}$ \sqrt{{(x+2)}^2+{(y-0)}^2}=2\sqrt{{(x-1)}^2+{(y-0)}^2}$$ \Rightarrow {(x+2)}^2+y^2=4\{{(x-1)}^2+y^2\}$$ \Rightarrow {(x+2)}^2-4{(x-1)}^2=3y^2$$ \Rightarrow x^2+4+4x-4x^2+8x-4=3y^2$$ \Rightarrow -3x^2+12x=3y^2$$ \Rightarrow x^2+y^2-4x=0$$ \Rightarrow {(x-2)}^2+y^2=4$\end{array}$