Rotation of Rectangular Hyperbola from Reduced to Standard Form

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Theorem

Let $\KK$ be a rectangular hyperbola embedded in a Cartesian plane in reduced form:

$x^2 - y^2 = a^2$

Let $\KK$ be rotated $45 \degrees$ clockwise about the origin.


Then $\KK$ is in standard form:

$x y = c^2$

where $c = \dfrac a {\sqrt 2}$.


Proof

Let $\tuple {x, y}$ denote an arbitrary point on $\KK$ before rotation.

Let $\tuple {x', y'}$ denote the image of $\tuple {x, y}$ after rotation.


Then:

\(\ds \begin {pmatrix} x' \\ y' \end {pmatrix}\) \(=\) \(\ds \begin {bmatrix} \map \cos {-45 \degrees} & -\map \sin {-45 \degrees} \\ \map \sin {-45 \degrees} & \map \cos {-45 \degrees} \end {bmatrix} \begin {pmatrix} x \\ y \end {pmatrix}\) Matrix Equation of Plane Rotation: clockwise is negative
\(\ds \) \(=\) \(\ds \begin {bmatrix} \map \cos {45 \degrees} & \map \sin {45 \degrees} \\ -\map \sin {45 \degrees} & \map \cos {45 \degrees} \end {bmatrix} \begin {pmatrix} x \\ y \end {pmatrix}\) Sine Function is Odd, Cosine Function is Even
\(\ds \) \(=\) \(\ds \begin {bmatrix} \dfrac {\sqrt 2} 2 & \dfrac {\sqrt 2} 2 \\ -\dfrac {\sqrt 2} 2 & \dfrac {\sqrt 2} 2 \end {bmatrix} \begin {pmatrix} x \\ y \end {pmatrix}\) Sine of $45 \degrees$, Cosine of $45 \degrees$
\(\ds \) \(=\) \(\ds \begin {bmatrix} \dfrac {\sqrt 2} 2 x + \dfrac {\sqrt 2} 2 y \\ -\dfrac {\sqrt 2} 2 x + \dfrac {\sqrt 2} 2 y \end {bmatrix}\) Definition of Matrix Product (Conventional)


Substituting for $x$ and $y$ in $x^2 - y^2 = a^2$:

\(\ds \paren {\dfrac {\sqrt 2} 2 x + \dfrac {\sqrt 2} 2 y}^2 - \paren {-\dfrac {\sqrt 2} 2 x + \dfrac {\sqrt 2} 2 y}\) \(=\) \(\ds a^2\)
\(\ds \leadsto \ \ \) \(\ds \paren {\paren {\dfrac {\sqrt 2} 2}^2 x^2 + 2 \times \paren {\dfrac {\sqrt 2} 2}^2 x y + \paren {\dfrac {\sqrt 2} 2}^2 y^2} - \paren {\paren {\dfrac {\sqrt 2} 2}^2 x^2 - 2 \times \paren {\dfrac {\sqrt 2} 2}^2 x y + \paren {\dfrac {\sqrt 2} 2}^2 y^2}\) \(=\) \(\ds a^2\) Square of Sum
\(\ds \leadsto \ \ \) \(\ds \paren {\dfrac 1 2 x^2 + x y + \dfrac 1 2 y^2} - \paren {\dfrac 1 2 x^2 - x y + \dfrac 1 2 y^2}\) \(=\) \(\ds a^2\) simplifying
\(\ds \leadsto \ \ \) \(\ds 2 x y\) \(=\) \(\ds a^2\) simplifying
\(\ds \leadsto \ \ \) \(\ds x y\) \(=\) \(\ds \dfrac {a^2} 2\)
\(\ds \) \(=\) \(\ds \paren {\dfrac a {\sqrt 2} }^2\)

Hence the result.

$\blacksquare$

Sources

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