Question 1137032
find an equation describing all points equidistant from the x-axis and the point ({{{0}}}, {{{2}}}). 

First, see if you can sketch a picture of what this curve ought to look like. 
For a point ({{{x}}}, {{{y}}}) that is on the curve, explain why {{{sqrt(y^2)= sqrt(x^2 - (y - 2)^2)}}}. 

note: 
{{{sqrt(y^2)}}} is the distance from  ({{{x}}}, {{{y}}}) to the x-axis
{{{sqrt(x^2 - (y - 2)^2)}}} is the distance from   ({{{x}}}, {{{y}}}) to the point ({{{0}}}, {{{2}}})

distances must be equal because is given:all points equidistant from the x-axis and the point ({{{0}}},{{{2}}})


Square both sides of this equation and solve for{{{ y}}}. Identify the curve. 

{{{y^2= x^2 - (y - 2)^2}}}

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

{{{y^2= x^2 - y^2 +4y-4}}}

{{{2y^2-4y= x^2  -4}}}......complete square for {{{y}}}

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

{{{2(y^2-2y+1^2)-2*1^2= x^2  -4}}}

{{{2(y-1)^2-2= x^2  -4}}}

{{{2(y-1)^2= x^2  -4+2}}}

{{{(y-1)^2= (x^2  -2)/2}}}

{{{y= sqrt(x^2/2  -1)+1}}}=>hyperbola


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(0,2,.12),locate(0,2,p(0,2)),
graph( 600, 600, -10, 10, -10, 10, sqrt(x^2/2  -1)+1, -sqrt(x^2/2  -1)+1)) }}}