The axis of a parabola is along the line y = x and the ?

Question

The axis of a parabola is along the line y = x and the distance of its vertex from origin is 2 and from focus is

2 *sqrt(2) . If vertex and focus both lie in first quadrant, then equation of the parabola is ?

Answer 1

And the cup ran away with the spoon....

Answer 2

First let's find the equation of a horizontal parabola in set of coordinate axes rotated 45° from the original set.  

The distance from the origin to the vertex is 2.  The vertex is:

(h,k) = (2,0)

The directed distance from the vertex to the focus is p.

p = 2√2

The equation of the parabola is:

4p(x' - 2) = (y' - 0)²

4(2√2)(x' - 2) = y'²

8√2(x' - 2) = y'²

________

x' = xcosθ + ysinθ

y' = -xsinθ + ycosθ

x' = xcos(π/4) + ysin(π/4) = x/√2 + y/√2

y' = -xsinθ + ycosθ = -x/√2 + y/√2

Now plug the values into the equation to get the equation in the original set of coordinates.

8√2(x' - 2) = y'²

8√2[(x/√2 + y/√2) - 2] = (-x/√2 + y/√2)²

8x + 8y - 16√2 = x²/2 - xy + y²/2

16x + 16y - 32√2 = x² - 2xy + y²

x² - 2xy + y² - 16x - 16y + 32√2 = 0