# 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, and otherwise, a non-degenerate conic section. This wiki page will give detailed information about the discriminant of a conic section.

## Nature of 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** (\(A \neq C\)). (For a real ellipse, \(\dfrac{\Delta}{a + b} < 0\).)

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

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 \[\begin{align} \Delta &=abc+2fgh-a f^2-bg^2-ch^2\\ &=5 \cdot 5 \cdot 2 +2\cdot 1 \cdot 2 \cdot 5- 5\cdot 1^2-5\cdot 2^2-2\cdot 5^2\\ &=50+20-5-20-50=-5 \ne 0. \end{align}\] 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 \(K?\)

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 \[\begin{align} \Delta &=abc+2fgh-a f^2-bg^2-ch^2\\ &=1 \cdot K \cdot 2 + 0-1\cdot 2^2-K\cdot 4^2-0\\ &=-4-14K=-4-14\cdot1 =-18 \ne 0. \end{align}\] 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?

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 \(B^2 – 4AC=\frac{-4}{m^2 n^1}< 0.\) 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 \[\begin{align} \Delta &=abc+2fgh-a f^2-bg^2-ch^2\\ &=\frac{1}{m^2} \cdot \frac{1}{n^2} \cdot (-4) + 0-0-0-0\\ &=\frac{-4}{m^2n^2} \ne 0. \end{align}\] This confirms \(\Delta\ne 0.\)

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

## Problem Solving

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

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

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 \[\begin{align} \Delta &=abc+2fgh-a f^2-bg^2-ch^2\\ &=3 \cdot 6 \cdot 1 +2\cdot 6 \cdot 3 \cdot 5- 3\cdot 6^2-6\cdot 3^2-1\cdot 5^2\\ &=18+180-108-54-25=11 \ne 0. \end{align}\] 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 \(J+K?\)

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 \[\begin{align} (x+3)^2+(y+7)^2-3^2-7^2+K&=0 \\ (x+3)^2+(y+7)^2&=58-K\\ &=r^2, \end{align}\] 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.

\[f(x , y) = nx^2 + \left(\dfrac{nm}{2}\right)xy + my^2 + 4x + 2y + 1 = 0\]

\[g(x , y) = nx^2 + (n + m)xy + my^2 + 4x + 2y + 1 = 0\]

\[h(x , y) = nx^2 + (n - m)xy + my^2 + 4x + 2y + 1 = 0\]

Let the graphs \(f, g, h\) be a parabola, hyperbola, and ellipse respectively.

If \(n\) and \(m\) are positive integers satisfying the constraints above, compute \(2n + m\).

## See Also

**Cite as:**Discriminant of a Conic Section.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/conics-discriminant/