Discriminant of a Conic Section

The general equation of a conic section is a second-degree equation in two independent variables (say $x,y$) which can be written as

$$f(x,y)=ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0.$$

There are several ways of classifying conic sections using the above general equation with the help of the discriminant $\Delta$ of this equation:

$$\Delta =\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}= abc+2fgh-a{ f }^{ 2 }-b{ g }^{ 2 }-c{ h }^{ 2 },$$

which describes the nature of the conic section. If $\Delta$ is zero, it represents a degenerate conic section; otherwise, it represents a non-degenerate conic section. This wiki page will give detailed information about the discriminant of a conic section.

Any second-degree curve equation can be written as

$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \qquad (1)$$

or

$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0, \qquad (2)$$

where $A,B,C,D,E,F,a,b,c,f,g,h \in \mathbb{R}.$

Now, let

$$\Delta =\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}= abc+2fgh-a{ f }^{ 2 }-b{ g }^{ 2 }-c{ h }^{ 2 },$$

then the type of conic section that the above equation represents can be found using the discriminant of the equation, which is given by $B^{2} - 4AC$ for $(1),$ or equivalently, $h^2-ab$ for $(2).$

The various conditions regarding the quadratic discriminant are as follows:

If $\Delta = 0:$

$\quad \bullet$ If $h^2-ab > 0,$ the equation represents two distinct real lines.
$\quad \bullet$ If $h^2-ab = 0,$ the equation represents parallel lines.
$\quad \bullet$ If $h^2-ab < 0,$ the equation represents non-real lines.

If $\Delta \neq 0:$

$\quad \bullet$ If $B^2-4AC > 0,$ it represents a hyperbola and a rectangular hyperbola $(A+C=0).$
$\quad \bullet$ If $B^2-4AC = 0,$ the equation represents a parabola.
$\quad \bullet$ If $B^2-4AC < 0,$ the equation represents a circle $(A=C, B= 0)$ or an ellipse $(A \neq C).$ $\big($For a real ellipse, $\frac{\Delta}{a + b} < 0.\big)$

What type of conic section does the following equation represent: $$5x^2 +10xy+5y^2 +4x+2y + 2 = 0?$$

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Equating the given equation with $(1)$ at the top of the page, we have $$A=5, B=10, C=5, D=4, E=2, F=2.$$ The discriminant for this equation is $B^2 - 4AC=10^2-4\cdot 5\cdot 5=0. \qquad \text{(i)}$

Rewriting the given equation in the form of $(2),$ we have $$5 \cdot x^2+2\cdot 5\cdot xy+5\cdot y^2+2\cdot 2\cdot x+ 2\cdot 1 \cdot y+2=0.$$ Equating this with $(2)$ gives $$a=5, b=5, c=2, f=1, g=2, h=5.$$ Now, checking $\Delta,$ we have $$\begin{aligned} \Delta &=abc+2fgh-a f^2-bg^2-ch^2\\ &=5 \cdot 5 \cdot 2 +2\cdot 1 \cdot 2 \cdot 5- 5\cdot 1^2-5\cdot 2^2-2\cdot 5^2\\ &=50+20-5-20-50=-5 \\&\ne 0. \end{aligned}$$ This confirms $\Delta\ne 0.$

Therefore, since $B^2 - 4AC=0$ from $\text{(i)},$ the equation represents a parabola. $_\square$

The equation $x^2 + Ky^2 + 8x + 4y + 2 = 0$ represents a circle. What is $K?$

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Equating the given equation with $(1)$ at the top of the page, we have $$A=1, B=0, C=K, D=8, E=4, F=2.$$ Thus, for the equation to represent a circle, the coefficients must simultaneously satisfy the discriminant condition $B^2 - 4AC< 0$ and also $A=C.$ The latter condition $A=C$ implies $K=1.$ Substituting this into $B^2 - 4AC< 0$ gives $0^2-4\cdot 1\cdot 1=-4<0,$ which is true.

Rewriting the given equation in the form of $(2),$ we have $$1 \cdot x^2+2\cdot 0\cdot xy+K\cdot y^2+2\cdot 4\cdot x+ 2\cdot 2y+2=0.$$ Equating this with $(2)$ gives $$a=1, b=K, c=2, f=2, g=4, h=0.$$ Now, checking $\Delta,$ we have $$\begin{aligned} \Delta &=abc+2fgh-a f^2-bg^2-ch^2\\ &=1 \cdot K \cdot 2 + 0-1\cdot 2^2-K\cdot 4^2-0\\ &=-4-14K=-4-14\cdot1 =-18 \\&\ne 0. \end{aligned}$$ This confirms $\Delta\ne 0.$

Therefore, $K=1.$ $_\square$

Note: The equation of the circle is $(x+4)^2+(y+2)^2=18.$

If $m$ and $n$ are nonzero real numbers, what is the condition on $m$ and $n$ such that the equation $$\frac{x^2}{m^2}+\frac{y^2}{n^2} = 4$$ represents an ellipse that is not a circle?

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Equating the given equation with $(1)$ at the top of the page, we have $$A=\frac{1}{m^2}, B=0, C=\frac{1}{n^2}, D=0, E=0, F=-4.$$ Thus, for the equation to represent an ellipse that is not a circle, the coefficients must simultaneously satisfy the discriminant condition $B^2 - 4AC< 0$ and also $A\ne C.$ The former condition is met because $B^2 - 4AC=\frac{-4}{m^2 n^2}< 0.$ The latter condition $A\ne C$ is satisfied if $m^2\ne n^2.$

Rewriting the given equation in the form of $(2),$ we have $$\frac{1}{m^2} \cdot x^2+2\cdot 0\cdot xy+\frac{1}{n^2}\cdot y^2+2\cdot 0\cdot x+ 2\cdot 0 \cdot y-4=0.$$ Equating this with $(2)$ gives $$a=\frac{1}{m^2}, b=\frac{1}{n^2}, c=-4, f=0, g=0, h=0.$$ Now, checking $\Delta,$ we have $$\begin{aligned} \Delta &=abc+2fgh-a f^2-bg^2-ch^2\\ &=\frac{1}{m^2} \cdot \frac{1}{n^2} \cdot (-4) + 0-0-0-0\\ &=\frac{-4}{m^2n^2}\\& \ne 0. \end{aligned}$$ This confirms $\Delta\ne 0.$

Therefore, the condition for the equation to represent an ellipse that is not a circle is $$m^2 \ne n^2. \ _\square$$

Check out the following examples based on the discriminant of a conic section.

What type of conic section does the following equation represent: $$3x^2 +10xy+6y^2 +6x+12y + 1 = 0?$$

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Equating the given equation with $(1)$ at the top of the page, we have $$A=3, B=10, C=6, D=6, E=12, F=1.$$ The discriminant for this equation is $B^2 - 4AC=10^2-4\cdot 3\cdot 6=28>0. \qquad \text{(i)}$

Rewriting the given equation in the form of $(2),$ we have $$3 \cdot x^2+2\cdot 5\cdot xy+6\cdot y^2+2\cdot 3\cdot x+ 2\cdot 6 \cdot y+1=0.$$ Equating this with $(2)$ gives $$a=3, b=6, c=1, f=6, g=3, h=5.$$ Now, checking $\Delta,$ we have $$\begin{aligned} \Delta &=abc+2fgh-a f^2-bg^2-ch^2\\ &=3 \cdot 6 \cdot 1 +2\cdot 6 \cdot 3 \cdot 5- 3\cdot 6^2-6\cdot 3^2-1\cdot 5^2\\ &=18+180-108-54-25=11 \\&\ne 0. \end{aligned}$$ This confirms $\Delta\ne 0.$

Therefore, since $B^2 - 4AC>0$ from $\text{(i)},$ the equation represents a hyperbola. $_\square$

Suppose that the equation $x^2+Jxy+y^2+6x+14y+K = 0$ represents a circle with radius $2.$ What is $J+K?$

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Equating the given equation with $(1)$ at the top of the page, we have $$A=1, B=J, C=1, D=6, E=14, F=K.$$ Thus, for the equation to represent a circle, the coefficients must simultaneously satisfy the discriminant condition $B^2 - 4AC< 0$ and also $A=C$ and $B=0.$ This last condition $B=0$ is met if $J=0.$ The second condition $A=C$ is already met because we are given $A=C=1.$ Then the the discriminant condition gives $$B^2 - 4AC=0^2-4\cdot 1\cdot 1=-4< 0,$$ which is true.

Thus, rewriting the given equation, we have $$\begin{aligned} (x+3)^2+(y+7)^2-3^2-7^2+K&=0 \\ (x+3)^2+(y+7)^2&=58-K\\ &=r^2, \end{aligned}$$ where $r=2$ is the radius of the circle. Hence, $K=58-2^2=54,$ implying $$J+K=0+54=54. \ _\square$$

Try the following problems.

• Equation of a Parabola

• Equation of an Ellipse

• Equation of a Hyperbola

• Equation of a Circle