Arun
Last Activity: 7 Years ago
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
