Question 1132430
The distance from {{{P}}}({{{x}}},{{{y}}}) to the line {{{x=-1}}} will taken from
the perpendicular distance as shortest which will be a horizontal distance since {{{x=-1}}} is a vertical line.

This distance will be the difference in the {{{x}}} coordinates
stated as a positive or {{{abs(x-(-1))}}} => {{{abs(x+1)}}}
 
Since we want the distance from {{{P}}}({{{x}}},{{{y}}})  to {{{K}}}({{{2}}},{{{5}}}) to be
twice the distance from ({{{x}}},{{{y}}}) to {{{x=-1}}} we have:
 
{{{sqrt((x-2)^2+(y-5)^2) = 2*abs(x+1)}}} ... square both sides

Note that since squaring {{{abs(x+1)}}} will automatically result in a positive we will no longer need the absolute value sign.
 
{{{(x-2)^2+(y-5)^2 = 4(x+1)^2}}}
 
{{{x^2-4x+4+y^2-10y+25 = 4(x^2+2x+1)}}}

{{{x^2-4x +y^2-10y+29 = 4x^2+8x+4}}}
 
{{{0 = 4x^2+8x+4-(x^2-4x +y^2-10y+29)}}}
 
{{{ 4x^2+8x+4-x^2+4x -y^2+10y-29=0}}}

{{{3x^2 + 12x - y^2 + 10y = 25}}}
 
Note this is becoming the equation of a hyperbola.

We want standard form {{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1 }}}:

We need to do some complete the square work here.
 
{{{3(x^2+4x) - (y^2-10y) = 25}}}

{{{3(x^2+4x+4) - (y^2-10y + 25) = 25 + 12 - 25}}}
 
{{{3(x+2)^2 - (y-5)^2 = 12}}}
 
{{{(x+2)^2/4 - (y-5)^2/12 = 1}}}
 
We have a hyperbola centered at ({{{h}}},{{{k}}}) or ({{{-2}}},{{{5}}}).=>the equation of the locus of {{{P}}}