User:John Wormell/haydos

Let (2,0) be S; let the point lying on the line x=8 closest to P be Q; its co-ordinates are (8,y) as PQ must be perpendicular to the line x=8 and must therefore have the equation y = y (the second y being the y in P(x,y))

so:

$PQ=8-x$

$PS={\sqrt {(x-2)^{2}+(y-0)^{2}}}$

$PS={\sqrt {(x-2)^{2}+y^{2}}}$

$PQ=2PS$

$8-x=2{\sqrt {(x-2)^{2}+y^{2}}}$

$8-x={\sqrt {4(x-2)^{2}+4y^{2}}}$

$(8-x)^{2}=4(x-2)^{2}+4y^{2}$

Expand that out and equate like terms and all that and it should come out.