Classifying and Diagonalizing a Quadratic Form (Conic Section): 04-BS-1, May 2015, Q7

Question 7: 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 $Q=-2x^2+12xy+7y^2$, set equal to $156$.

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

Approach. Write $Q=\mathbf x^TM\mathbf x$ with $M$ symmetric ($M_{12}=M_{21}=$ half the $xy$-coefficient), find its eigenvalues (these become $a,b$ in the rotated frame), and classify by their signs: same sign $\to$ ellipse, opposite signs $\to$ hyperbola.

Symmetric matrix of the form. $Q=\mathbf x^TM\mathbf x$ with $$M=\begin{pmatrix}-2&6\\6&7\end{pmatrix}$$ (the off-diagonal entries split the $12xy$ coefficient evenly).
Eigenvalues of $M$. $\det(M-\lambda I)=(-2-\lambda)(7-\lambda)-36=\lambda^2-5\lambda-50=0\Rightarrow\lambda=\dfrac{5\pm\sqrt{25+200}}{2}=\dfrac{5\pm15}{2}$, so $$\lambda_1=10,\qquad\lambda_2=-5.$$
Classify the conic. The eigenvalues have opposite signs ($10$ and $-5$), so $Q=156$ is a hyperbola.
Principal-axis directions (for reference). For $\lambda_1=10$: $(M-10I)\mathbf v=0\Rightarrow-12x+6y=0\Rightarrow\mathbf v_1\parallel(1,2)$. For $\lambda_2=-5$: $(M+5I)\mathbf v=0\Rightarrow3x+6y=0\Rightarrow\mathbf v_2\parallel(-2,1)$ — orthogonal to $\mathbf v_1$, as required ($(1,2)\cdot(-2,1)=0$).
Principal-axis form. In the rotated coordinates $u,v$ aligned with $\mathbf v_1,\mathbf v_2$, the quadratic form diagonalizes to the eigenvalues: $$\boxed{Q=10u^2-5v^2=156}.$$

Quantity	Result
Eigenvalues of $M$	$\lambda_1=10,\ \lambda_2=-5$
Conic type	Hyperbola (opposite-sign eigenvalues)
Principal-axis directions	$(1,2)$ and $(-2,1)$ (orthogonal)
Principal-axis form	$Q=10u^2-5v^2=156$

Concept & Theory

Topic. The principal axis theorem: diagonalizing a quadratic form via the eigenvalues/eigenvectors of its symmetric matrix.

Key relations. Any quadratic form $Q=ax^2+bxy+cy^2$ can be written $\mathbf x^TM\mathbf x$ with $M=\begin{pmatrix}a&b/2\\b/2&c\end{pmatrix}$ symmetric; since $M$ is symmetric it has real eigenvalues $\lambda_1,\lambda_2$ with orthogonal eigenvectors, and in the coordinate system aligned with those eigenvectors, $Q=\lambda_1u^2+\lambda_2v^2$ exactly — no cross term.

Why this works. The eigenvectors of a symmetric matrix are orthogonal (spectral theorem), so rotating the coordinate axes to align with them is a genuine rigid rotation (preserves distances/angles) that simultaneously eliminates the cross term — this is precisely why the $xy$-term in a conic's equation can always be removed by a rotation, never a shear.

Common pitfall. Using the full off-diagonal coefficient ($6$) instead of half the $xy$-coefficient when building $M$ from $Q=-2x^2+12xy+7y^2$ — the matrix entry is $12/2=6$, not $12$, since $\mathbf x^TM\mathbf x$ counts each off-diagonal contribution twice.

Source. Strang, Introduction to Linear Algebra, Ch. 6 (§6.4 Quadratic Forms and the Principal Axis Theorem).

Question 7 of 8: Classifying and Diagonalizing a Quadratic Form (Conic Section)

Question 7: Classifying and Diagonalizing a Quadratic Form (Conic Section) (20 marks)