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.