What is the equation for the parabola tilted 45 degrees?

Question

What function y = f(x) describes the parabola y = x² that has been tilted 45°, so that f(0) = 0, and x = f(y). That is, the function y = f(x) is its own inverse?

TheMathematician, I'm using a looser definition of "function", whereby if you have the expression, you can (or some math software can) actually plot it. As Alexander has pointed out, the resulting quadratic expression is "easily solvable". And, yes, it is it's own true inverse.

Answer 1

I am not sure what you mean by saying "y = f(x) is its own inverse." If you tilt a parabola by 45 degrees, then it will no longer be a function (it fails the vertical line test), so it is not possible to solve its equation for y.

However, I can tell you the equation of a parabola that has been tilted 45° clockwise. (This is closest to the spirit of the problem, because the graph will then be symmetric about the line y = x, a property which is shared by the graphs of functions which are their own inverses.)

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.

This cannot be solved for y (nor would we expect it could be, because the graph of the parabola is not a function).

EDIT: Oops. Of course, as the answerer after me points out, the equation can be solved for y by viewing it as a quadratic equation in y whose coefficients are polynomials in x:

y^2 - (sqrt(2) + 2x)y + (x^2 - x * sqrt(2)) = 0

Use of the quadratic formula yields the two pieces

y = [(sqrt(2) + 2x) +/- sqrt(4x^2 + 2 + 4 sqrt(2) x - 4x^2 + 4 sqrt(2) x)] / 2

or

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

Answer 2

Tilted Parabola

Answer 3

this equation copied from the answer above

x^2 - 2xy + y^2 - x * sqrt(2) - y * sqrt(2) = 0

is quite simply quadratic equation with respect to y, and can be very easily solved