Homogeneous equation of degree two

A homogeneous equation of degree two in $x$ and $y$ has the general form $ax^2 + 2hxy + by^2 = 0$. This section shows exactly when such an equation is the combined equation of a pair of lines through the origin, and how to recover the slopes of those lines.

Theorem 4.2. The combined equation of a pair of lines passing through the origin is a homogeneous equation of degree two in $x$ and $y$.

Proof. Let $a_1x + b_1y = 0$ and $a_2x + b_2y = 0$ be any two lines through the origin.

Their combined equation is $(a_1x+b_1y)(a_2x+b_2y) = 0$, which expands to $a_1a_2x^2 + (a_1b_2+a_2b_1)xy + b_1b_2y^2 = 0$. Writing $a = a_1a_2$, $2h = a_1b_2 + a_2b_1$, and $b = b_1b_2$, this is exactly $ax^2 + 2hxy + by^2 = 0$, a homogeneous equation of degree two (every term -- $x^2$, $xy$, $y^2$ -- has degree two). $\blacksquare$

The converse is not automatic: every homogeneous equation of degree two need not represent a real pair of lines. For instance $x^2 + y^2 = 0$ is homogeneous of degree two, but its only real solution is the single point $(0,0)$, not two lines. The next theorem pins down exactly when a homogeneous equation does represent a genuine pair of lines.

Theorem (the $h^2 - ab$ test). The homogeneous equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of lines through the origin if and only if $h^2 - ab \geq 0$.

Proof. Consider $ax^2 + 2hxy + by^2 = 0 \quad \ldots (1)$, and split into the two exhaustive cases $b = 0$ and $b \neq 0$.

Case 1 ($b=0$). Equation (1) becomes $ax^2 + 2hxy = 0$, i.e. $x(ax + 2hy) = 0$, which is exactly the combined equation of the two lines $x = 0$ and $ax + 2hy = 0$ -- both of which visibly pass through the origin.

Case 2 ($b \neq 0$). Multiply (1) by $b$: $abx^2 + 2hbxy + b^2y^2 = 0$, i.e. $b^2y^2 + 2hbxy = -abx^2$. Completing the square on the left by adding $h^2x^2$ to both sides, $b^2y^2 + 2hbxy + h^2x^2 = h^2x^2 - abx^2$, i.e. $(by+hx)^2 = (h^2-ab)x^2$. Provided $h^2 - ab \geq 0$, the right side is a perfect square, so $(by+hx)^2 - (h^2-ab)x^2 = 0$ factors as a difference of squares: $\Big[(h+\sqrt{h^2-ab})x + by\Big]\Big[(h-\sqrt{h^2-ab})x + by\Big] = 0$. This is the combined equation of the two lines $(h+\sqrt{h^2-ab})x + by = 0$ and $(h-\sqrt{h^2-ab})x + by = 0$, and since $b \neq 0$ we may write these as $y = m_1x$ and $y = m_2x$ with $m_1 = \dfrac{-h-\sqrt{h^2-ab}}{b}$ and $m_2 = \dfrac{-h+\sqrt{h^2-ab}}{b}$ -- again visibly through the origin. $\blacksquare$

Remarks. (1) If $h^2-ab > 0$, the two lines are distinct. (2) If $h^2-ab = 0$, the two lines are coincident (they collapse onto the same line). (3) If $h^2-ab < 0$, equation (1) has no real solution besides the origin, so it does not represent a genuine pair of lines. (4) If $b=0$, one line is the $Y$-axis (undefined slope) and the other has slope $-\dfrac{a}{2h}$ (provided $h \neq 0$). (5) If $h^2-ab \geq 0$ and $b \neq 0$, the two slopes are $m_1 = \dfrac{-h-\sqrt{h^2-ab}}{b}$ and $m_2 = \dfrac{-h+\sqrt{h^2-ab}}{b}$, so their sum is $m_1+m_2 = \dfrac{-2h}{b}$ and their product is $m_1m_2 = \dfrac{a}{b}$.

Because $m_1,m_2$ are roots of a quadratic with that sum and product, the quadratic $m^2 - (m_1+m_2)m + m_1m_2 = 0$ becomes, after multiplying through by $b$, $bm^2 + 2hm + a = 0 \quad \ldots (2)$. Equation (2) is called the auxiliary equation of (1); its two roots are precisely the slopes of the two lines represented by (1). This is the standard tool used throughout the chapter to extract slopes from a homogeneous pair.

Worked Examples

Example 1. Show that the lines represented by $x^2 - 2xy - 3y^2 = 0$ are distinct.

Here $a=1, h=-1, b=-3$, so $h^2-ab = (-1)^2-(1)(-3) = 1+3 = 4 > 0$. Since $h^2-ab>0$, the two lines are distinct.

Example 2. Show that the lines represented by $x^2 - 6xy + 9y^2 = 0$ are coincident.

Here $a=1,h=-3,b=9$, so $h^2-ab = 9-9=0$. Since $h^2-ab=0$, the lines are coincident.

Example 3. Find the sum and product of the slopes of the lines represented by $x^2+4xy-7y^2=0$.

Here $a=1,h=2,b=-7$. Using the sum/product formulas, $m_1+m_2 = \dfrac{-2h}{b} = \dfrac{-4}{-7} = \dfrac{4}{7}$ and $m_1m_2 = \dfrac{a}{b} = \dfrac{1}{-7} = -\dfrac{1}{7}$.

Example 4. Find the separate equations of the lines represented by (i) $x^2-4y^2=0$, (ii) $3x^2-7xy+4y^2=0$, (iii) $x^2+2xy-y^2=0$, (iv) $5x^2-3y^2=0$.

1. $x^2-4y^2=(x-2y)(x+2y)=0$, so the lines are $x-2y=0$ and $x+2y=0$.
2. $3x^2-7xy+4y^2$: split the middle term as $-3xy-4xy$: $3x^2-3xy-4xy+4y^2 = 3x(x-y)-4y(x-y) = (x-y)(3x-4y)=0$, giving lines $x-y=0$ and $3x-4y=0$.
3. $x^2+2xy-y^2=0$ does not factor over the rationals, so use the auxiliary equation $bm^2+2hm+a=0$ with $a=1,h=1,b=-1$: $-m^2+2m+1=0$, i.e. $m^2-2m-1=0$, giving $m = \dfrac{2\pm\sqrt{8}}{2} = 1\pm\sqrt2$. The lines are $y=(1+\sqrt2)x$ and $y=(1-\sqrt2)x$, i.e. $(1+\sqrt2)x-y=0$ and $(1-\sqrt2)x-y=0$.
4. $5x^2-3y^2 = (\sqrt5x)^2-(\sqrt3y)^2=(\sqrt5x-\sqrt3y)(\sqrt5x+\sqrt3y)=0$, giving lines $\sqrt5x-\sqrt3y=0$ and $\sqrt5x+\sqrt3y=0$. Example 5. Find $k$ if $2x+y=0$ is one of the lines represented by $3x^2+kxy+2y^2=0$. The slope of $2x+y=0$ is $-2$. Since this line is one of the pair, $-2$ must be a root of the auxiliary equation $2m^2+km+3=0$ (here $a=3,h=k/2,b=2$, so $bm^2+2hm+a = 2m^2+km+3$). Substituting $m=-2$: $2(4)+k(-2)+3=0 \Rightarrow 8-2k+3=0 \Rightarrow 2k=11 \Rightarrow k=\dfrac{11}{2}$. Alternatively, since every point of $2x+y=0$ satisfies the combined equation, take the point $(1,-2)$ on that line and substitute into $3x^2+kxy+2y^2=0$: $3(1)+k(1)(-2)+2(4)=0 \Rightarrow 3-2k+8=0 \Rightarrow 2k=11 \Rightarrow k=\dfrac{11}{2}$, confirming the same value. Example 6. Find the condition that $3x-2y=0$ coincides with one of the lines represented by $ax^2+2hxy+by^2=0$. …