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Question 715758: equation of a tilted parabola
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! First, notice that the equation of the parabola y = x^2 can be parametrized by x = t, y = t^2, as t goes from -infinity to infinity; or, as a column vector,
[x] = [t]
[y] = [t^2].
To rotate the graph of the parabola about the origin, we rotate each point individually. Rotation clockwise by 45 degrees is a linear transformation; the transformation sends the point (1, 0) to (1/sqrt(2), -1/sqrt(2)), and it sends the point (0, 1) to (1/sqrt(2), sqrt(2)). So the standard matrix for the linear transformation is
[1/sqrt(2) 1/sqrt(2)]
[-1/sqrt(2) 1/sqrt(2)]
Thus, if we apply this linear transformation to a point (t, t^2) on the graph of the parabola, we get
[1/sqrt(2) 1/sqrt(2)][t]
[-1/sqrt(2) 1/sqrt(2)][t^2]
Which gives
[(t^2 + t)/sqrt(2)]
[(t^2 - t)/sqrt(2)].
So, as t goes from -infinity to infinity, this is a parametrization of the graph of the rotated parabola. We must now convert back to x and y.
We have
[x] = [(t^2 + t)/sqrt(2)]
[y] = [(t^2 - t)/sqrt(2)]
By adding and subtracting x and y, we find
x - y = t * sqrt(2)
x + y = t^2 * sqrt(2)
Solving the first equation for t, we get t = (x - y) / sqrt(2). Plug this into the second equation:
x + y = ((x - y) / sqrt(2))^2 * sqrt(2)
x + y = ((x^2 - 2xy + y^2) / 2) * sqrt(2)
x + y = (x^2 - 2xy + y^2) / sqrt(2)
x * sqrt(2) + y * sqrt(2) = x^2 - 2xy + y^2
x^2 - 2xy + y^2 - x * sqrt(2) - y * sqrt(2) = 0.
So an equation for the parabola y = x^2 rotated clockwise by 45 degrees is
x^2 - 2xy + y^2 - x * sqrt(2) - y * sqrt(2) = 0.
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Copied from somewhere, I lost the link to the site.
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