Question 2 of 8: Classifying and Diagonalizing a Quadratic Form (Conic Section)

Question 2: Classifying and Diagonalizing a Quadratic Form (Conic Section) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Quadratic form $Q=-2x^2+12xy+7y^2=156$.

Find. The conic type, and the principal-axis form $Q=au^2+bv^2$.

Approach. Write $Q=\mathbf{x}^TA\mathbf{x}$ with symmetric matrix $A$, find $A$'s eigenvalues (these become $a,b$) and orthonormal eigenvectors (these define the rotated $u,v$ axes), then classify by the signs of the eigenvalues.

1. Build the symmetric matrix. Writing $Q=ax^2+2bxy+cy^2$ with $a=-2,\ 2b=12\Rightarrow b=6,\ c=7$, $$A=\begin{pmatrix}-2&6\\6&7\end{pmatrix}.$$
2. Eigenvalues. $$\det(A-\lambda I)=(-2-\lambda)(7-\lambda)-36=\lambda^2-5\lambda-50=(\lambda-10)(\lambda+5)=0 \;\Rightarrow\; \boxed{\lambda_1=10,\ \lambda_2=-5}.$$
3. Eigenvectors (principal axes). For $\lambda_1=10$: $(A-10I)v=0\Rightarrow\begin{pmatrix}-12&6\\6&-3\end{pmatrix}v=0\Rightarrow v_1=2v_2$, direction $(1,2)$, unit vector $\hat u=\tfrac1{\sqrt5}(1,2)$. For $\lambda_2=-5$: $(A+5I)v=0\Rightarrow\begin{pmatrix}3&6\\6&12\end{pmatrix}v=0\Rightarrow v_1=-2v_2$, direction $(-2,1)$, unit vector $\hat v=\tfrac1{\sqrt5}(-2,1)$ (orthogonal to $\hat u$, as expected for a symmetric matrix).
4. Assemble the principal-axis form and classify. $$Q=10u^2-5v^2=156.$$ The eigenvalues have opposite signs, so the conic is a hyperbola. Dividing through, $\dfrac{u^2}{15.6}-\dfrac{v^2}{31.2}=1$.

$$\boxed{Q=10u^2-5v^2=156\ \text{(hyperbola)},\quad u\ \text{along}\ (1,2)/\sqrt5,\quad v\ \text{along}\ (-2,1)/\sqrt5}$$