Question

Under a rotation of axes, a parabola can become a hyperbola.
Is the statement true? Give reason for your answer, either with a short proof or a counterexample.

Answer 1

Answer on Question #78512 – Math – Analytic Geometry

Question

Under a rotation of axes, a parabola can become a hyperbola. Is the statement true? Give reason for your answer, either with a short proof or a counterexample.

Solution

The general equation of the second-order curve is given by

$$a_{11}x^2 + 2a_{12}xy + a_{22}y^2 + 2a_{31}x + 2a_{23}y + a_{33} = 0,$$

where $a_{ii}$ are some coefficients. The class of the curve (ellipse, hyperbola, parabola) is defined by the following determinant

$$\delta = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}$$

If $\delta > 0$ then the curve is ellipse, if $\delta = 0$, then parabola, else hyperbola.

First three members of the equation may be rewritten in the following way

$$Q(x,y) = \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$

Rotation of axes is given by

$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x' \\ y' \end{pmatrix},$$

where $x', y'$ are new coordinates.

Substituting it into $Q(x,y)$ we obtain

$$Q(x,y) = \begin{pmatrix} x' & y' \end{pmatrix} \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}^T \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x' \\ y' \end{pmatrix}$$

Thus,

$$\delta' = \det \left\{ \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}^T \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \right\} =$$ $$= \det \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}^T \det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \det \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} = |\sin^2 \theta + \cos^2 \theta| = 1 =$$ $$= \det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \delta$$

This means that under such transformation $\delta$ doesn't change, thus the curve can't change its class. So the statement is false.

Answer: false.

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