Chapter 8.4, Problem 67E

Solution

To prove that the discriminant is invariant under axes rotation.

Given information:

The equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

Proof:

The discriminant of the equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ is $B^2-4AC$ .

Substitute $x=u\cos \theta -v\sin \theta$ and $y=u\sin \theta +v\cos \theta$ in the equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ to find the transformed equation after rotation by an angle $\theta$ .

  $\begin{array}{c}A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0\\ \left(\begin{array}{l}A{(}^u+B(u\cos \theta -v\sin \theta )(u\sin \theta +v\cos \theta )+C{(}^u\\ +D(u\cos \theta -v\sin \theta )+E(u\sin \theta +v\cos \theta )+F\end{array}\right)=0\\ \left(\begin{array}{l}{u}^{2}A{\cos }^2\theta -uv2A\cos \theta \sin \theta +v^{2}A{\sin }^2\theta +u^{2}B\cos \theta \sin \theta +uvB{\cos }^2\theta -uvB{\sin }^2\theta \\ -v^{2}B\cos \theta \sin \theta +u^{2}C{\sin }^2\theta +uv2C\sin \theta \cos \theta +v^{2}C\cos \theta +uD\cos \theta \\ -vD\sin \theta +uE\sin \theta +vE\cos \theta +F\end{array}\right)=0\\ \left(\begin{array}{l}{u}^2\left(A{\cos }^2\theta +B\cos \theta \sin \theta +C{\sin }^2\theta \right)+uv\left(B{\cos }^2\theta -B{\sin }^2\theta +(C-A)2\sin \theta \cos \theta \right)\\ +v^2\left(C{\cos }^2\theta -B\cos \theta \sin \theta +A{\sin }^2\theta \right)+u(D\cos \theta -E\sin \theta )+v(E\cos \theta -D\sin \theta )+F\end{array}\right)=0\\ \left(\begin{array}{l}{u}^2\left(A{\cos }^2\theta +B\cos \theta \sin \theta +C{\sin }^2\theta \right)+uv\left(B\cos 2\theta +(C-A)\sin 2\theta \right)\\ +v^2\left(C{\cos }^2\theta -B\cos \theta \sin \theta +A{\sin }^2\theta \right)+u(D\cos \theta +E\sin \theta )+v(E\cos \theta -D\sin \theta )+F\end{array}\right)=0\end{array}$

Here, $A\text{'}=A{\cos }^2\theta +B\cos \theta \sin \theta +C{\sin }^2\theta$ , $B\text{'}=B\cos 2\theta +(C-A)\sin 2\theta$ and $C\text{'}=C{\cos }^2\theta -B\cos \theta \sin \theta +{\sin }^2\theta$ .

Now, find $B{\text{'}}^2-4A\text{'}C\text{'}$ .

  $\begin{array}{c}B{\text{'}}^2-4A\text{'}C\text{'}={\left(B\cos 2\theta +(C-A)\sin 2\theta \right)}^2-4\left(A{\cos }^2\theta +B\cos \theta \sin \theta +C{\sin }^2\theta \right)\left(C{\cos }^2\theta -B\cos \theta \sin \theta +A{\sin }^2\theta \right)\\ =\left(\begin{array}{l}{B}^2{\cos }^2(2\theta )+BC\sin (4\theta )\\ -BA\sin (4\theta )+{(}^{\sin }\end{array}\right)-\left(\begin{array}{l}-4CA{\cos }^4(\theta )+4BA{\cos }^3(\theta )\sin (\theta )-4A^2{\cos }^2(\theta ){\sin }^2(\theta )\\ -4BC{\cos }^3(\theta )\sin (\theta )+4B^2{\cos }^2(\theta ){\sin }^2(\theta )-4BA{\sin }^3(\theta )\cos (\theta )\\ -4C^2{\sin }^2(\theta ){\cos }^2(\theta )+4BC{\sin }^3(\theta )\cos (\theta )-4CA{\sin }^4(\theta )\end{array}\right)\\ =B^2-4AC\end{array}$

This shows that the discriminant is invariant under axes rotation.