why write a conic in this form?

for any real A, B, C, D, E, and F, a conic may be expressed as

$$Ax^2 + Bxy +Cy^2+Dx+Ey+F=0$$

why wouldn't it be more natural to write the conic in standard form with some kind of center $(h, k)$ and all of its other features encoded into $a$ and $b$?

Why go through the trouble of teaching students to go from the general form to the standard form? Expanding it out from standard form to general form is also a very lengthy process but why might someone want that?

(I don't have an issue with completing the square)

Is it because of the study of second-order differential equations?


  • 4
    $\begingroup$ Write the equation of a cone. Write the equation of a plane. Have them intersect. Simplify the result => you get the general formula of a comic section. $\endgroup$
    – Dominique
    Mar 12 at 13:07
  • 4
    $\begingroup$ Your question could be improved if you started with your answer to to your own question, but in the context of the general form of the equation of a line in the plane: $ax+by+c=0$. Why not just teach the form $y=mx+b$ or maybe $y-y_0=m(x-x_0)$. Similarly for the general form of the equation of a plane in space. $\endgroup$
    – user52817
    Mar 12 at 16:11
  • 2
    $\begingroup$ @Dominique I'm still trying to work out whether “comic section” was a deliberate pun… $\endgroup$
    – gidds
    Mar 13 at 21:16
  • 1
    $\begingroup$ @gidds: very deliberate ;-) $\endgroup$
    – Dominique
    Mar 14 at 11:04

6 Answers 6


What general form do you propose instead? You seem to think there is a better one, but I think probably the one you are thinking of is not able to encode all the conics.

I'm guessing based on your question, but you are probably looking for something like

$\pm \frac{(x-h)^2}{a^2} \pm \frac{(y-k)^2}{b^2} = 1$

as a candidate general form instead.

However, many simple conic sections cannot be written that way, like for example the hyperbola

$xy = 1$

has no $h, k, a, b$ that would express it in the proposed form.

A graph of <span class=$xy=1$" />

  • $\begingroup$ that makes a lot of sense, thanks! $\endgroup$
    – Lenny
    Mar 14 at 1:50

The point is to start with the general quadratic equation in two variables, and show that the so-called "non-degenerate" cases lead to the three types of curves (ellipse, parabola, hyperbola). Note that not all solutions of the general quadratic equation are literally conic sections, i.e., not all are identifiable with the intersection of some plane with a standard cone in 3-space. For example, a pair of parallel lines can arise as the solution of a quadratic equation, but cannot arise literally as a planar section of a cone. The same goes for the empty set (a literal conic section is always nonempty).


Because the general equation and the standard equation may arise in different contexts

Context 1: no a priori coordinate system: for example, you are given a conic section either as a locus of points (e.g. foci and length of axes for an ellipse) or a real-life problem (parabolic bridge, satellite dish, etc) that you want to study.

In that case, you are free to set a coordinate system (with perpendicular axes and the same unit on each axis) the way you prefer. In that case, it is a good choice to take one axis parallel (or perpendicular) to the major/transverse/symmetry axis of the conic. In that case, the formulas for distances and a bit of algebra lead to the standard equations.

(Good idea to also place the origin at the center of the ellipse/hyperbola or the vertex of the parabola).

Context 2: a coordinate system is given: For example, a curve is given by an equation in the plane, and you need to identify it, or study it, or sketch it. In that case, if the equation is equivalent to a quadratic equation, there is no reason for this equation to be in standard form. We need knowledge on the general equation of conic sections to deal with these situations.


  1. The graph of functions $f(x)=\frac{ax+b}{cx+d}$ is a hyperbola when $c\neq 0$ and $ad-bc\neq 0$. I introduce the rotation of axes by considering the graph of $y=\frac{1}{x}$ ($\iff xy=1$) and discussing with students how to show it is really a hyperbola.
  2. The graph of $\sqrt{x}+\sqrt{y}=1$ is part of a parabola.
  • $\begingroup$ I believe I have been focusing much on the first scenario $\endgroup$
    – Lenny
    Mar 13 at 10:45
  • 2
    $\begingroup$ @Lenny: Scenario 1 is probably the most common. Mot calculus problems where one has to compute a length, area, volume, etc related to a conic section will probably use a conic section in standard form $\endgroup$
    – Taladris
    Mar 13 at 11:13

Here is one way to think about the implications of the general form of a conic: $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$.

We have six parameters ($A,B, C, D, E, F$) but since the right hand side of the equation is 0, we effectively have just five parameters. (Think about forcing one of the parameters like $A$ to be equal to 1.)

From this point of view, we can expect that the space of conics in the plane is five-dimensional.

Now let's think about the celebrated theorem of Pascal, which tells us that 5 points in the plane determine a conic! So you give me 5 points, and I can find a unique conic that passes through all five points...sort of like two points determine a line. This is really a theorem. It is not obvious!

Now let's connect our heuristic counting with Pascal's theorem. If we are given one point $(x_1,y_1)$ then this imposes one linear constraint on our five variables: $$Ax_1^2+Bx_1y_1+Cy_1^2+Dx_1+Ey_1+F=0$$

Then we impose a second constraint by requiring the conic to pass through $(x_2,y_2)$, etc. Our counting is this:

0 constraints: 5 degrees of freedom (all conics)

1 constraint (force 1 point): 4 degrees of freedom

2 constraints (force 2 points): 3 degrees of freedom

3 constraints (force 3 points): 2 degrees of freedom (called a "net of conics")

4 constraints (force 4 points): 1 degree of freedom (called a "pencil of conics")

5 constraints (force 5 points): 0 degrees of freedom, so a unique conic.

We see that our intuitive expectation that the space of conics is 5-dimensional based on the general form of a conic is consistent with Pascal's Theorem. Perhaps this is why Pascal suspected the theorem in the first place.

Given five points $(x_1,y_1),\ldots,(x_5,y_5)$, we could solve the system of five linear equations in six unknowns to actually determine $A, B, C, D, E, F$ up to a scalar multiple. We could then look at $B^2-4AC$ to determine the type of the conic that passes through the five given points.

  • $\begingroup$ I wish you had a name, so I could remember you. This is beautiful. $\endgroup$
    – Sue VanHattum
    Mar 15 at 18:26
  • $\begingroup$ I'd briefly remark that this is just one strong example that illustrates the medium to long-term pedagogical benefits of understanding conics (quadratics) in the parameterized form. This Pascal's theorem flavor of thinking crops subsequently not just in algebraic geometry, but also as a beginning example in machine learning via least-squares fitting of polynomials to data. The scheme of the argument generalizes to higher degrees and more variables. This is the "standard form" that generalizes, with centered and other forms being special reparameterizations. $\endgroup$
    – user176372
    Mar 21 at 16:25

A major advantage to writing conic sections in standard form is the ease with which you can apply techniques like Joachimsthal's notation to solve problems such as finding tangents. In fact this is selling Joachimsthal short: you can do a lot more than finding tangents, and if you've not heard of it then it's worth reading this article in Cut-The-Knot and this more technical exposition by Wilson Stothers. This area used to be a major topic in the British A-level syllabus, and this Math SE question suggests it is still taught at high school level in the Netherlands. Even if this is a bag of tricks you're unfamiliar with, the standard form of conics also makes it straightforward to apply implicit differentiation. In fact in current British A-level textbooks, the chapter on implicit differentiation is where you'd come across finding tangents and normals to conic sections. I'll quote you some of the standard results about circles given in a 1980s A-level textbook, which still gives a sense of the "old skool" way students were taught to do things without resorting to calculus: but even by this time, many conics topics had been removed from the syllabus.

  • The general equation of a circle is $$x^2 + y^2 + 2gx + 2fy + c = 0$$ The circle has centre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$.
  • The equation of the circle with $AB$ as diameter where $A$ is $(x_1, y_1)$ and $B$ is $(x_2, y_2)$ is given by $$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$$
  • To find the equation of a circle passing through three points, substitute the coordinates of each point in turn into the general equation, solve the three simultaneous equations for the values of $g$, $f$ and $c$, then substitute these into the general equation.
  • The equation of the tangent at $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $$xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$$ Note the relation between the equations: $x^2 \to xx_1$, $y^2 \to yy_1$, $2x \to (x+x_1)$, $2y \to (y+y_1)$.
  • The length of the tangent from $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $$\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$$
  • The equation of any circle through the intersection of two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ is given by an equation of the form $$(x^2 + y^2 + 2g_1x + 2f_1y + c_1) + \lambda(x^2 + y^2 + 2g_2x + 2f_2y + c_2) = 0$$ where $\lambda$ is a constant that is usually found from the given conditions. If $\lambda = -1$, the equation reduces to a straight line. This is the common chord of the two circles and is known as their radical axis.

The above makes heavy use of the standard form of the equation for a circle, but a lot of that material will generalise to other conics. For example, using the notation at Cut-The-Knot, say we have the general conic

$$ Ax^2 + 2Bxy + Cy^2 + 2Fx + 2Gy + H=0 \tag{1}$$

One advantage of this form is we can (as Stothers does) write our equation succinctly in matrix form as

$$ \begin{bmatrix}x & y & 1 \end{bmatrix} \begin{bmatrix}A & B & F \\ B & C & G \\ F & G & H \end{bmatrix} \begin{bmatrix}x \\ y \\ 1 \end{bmatrix} = 0 \tag{2}$$

or simply $\mathbf{x'Mx} = 0$ where $\mathbf{M}$ is a symmetric 3-by-3 matrix. If the coordinates vector $\mathbf{x}$ looks suspiciously like it should have a $z$ but we have for some reason set $z=1$, then welcome to the realm of homogeneous coordinates and projective geometry — this setting unifies the treatment of conics and makes sense of many of their features, such as tangents and asymptotes, and "ideal points" at infinity. This matrix form gives us a test for whether a conic is degenerate: this occurs if and only if $\det \mathbf{M} = 0$. The determinant of the top left 2-by-2 sub-matrix is $\Delta = AC - B^2$ and gives a test to classify conics as ellipses, parabolas or hyperbolas depending on whether $\Delta > 0$ (ellipse, or in the degenerate case, a single point), $\Delta = 0$ (parabola, or if real solutions exist in the degenerate case, two possibly coincident parallel lines) or $\Delta < 0$ (hyperbola, or in the degenerate case, two intersecting lines).

Writing the conic as $(1)$ or $(2)$ lets us generalise the technique for finding the equation of a circle from three points, to find the equation of a general conic if given five points. (There are six coefficients and only five simultaneous equations, but solutions for the coefficients for the general conic can differ by a multiplicative constant. For the standard equation of a circle we fixed the $x^2$ as $y^2$ coefficients to one, so this issue didn't arise, but for a general conic we aren't guaranteed any particular coefficient is non-zero, so can't necessarily pick one to fix in this way.)

The textbook extract showed how to find the pencil of circles passing through the intersection points of two given circles, by taking the weighted sum of the circle's equations (in standard form). The textbook weighted the first equation by one and second by $\lambda$, so $\lambda = -1$ gave the degenerate case of the radical axis, $\lambda = 0$ recovers the first circle, but the second circle is only recovered in the limit $\lambda \to \infty$. To keep things real, you may prefer to weight the first equation by $\lambda$ and the second by $\mu$, not both zero. Then $(\lambda, \mu)=(1,0)$ recovers the first circle, $(0,1)$ recovers the second circle, and $(1,-1)$ gives the radical axis. This doesn't really give us an extra degree of freedom since multiplying $\lambda$ and $\mu$ by a common factor produces the same circle: all that matters is the ratio $\lambda:\mu$ and we are once again in the realm of homogeneous coordinates. (What the textbook denoted by $\lambda$ is in our notation the ratio $\mu/\lambda$, which is often what we want to work with anyway.) We can similarly find a pencil of conics through the intersection points of two given conics $\mathbf{x'Mx} = 0$ and $\mathbf{x'Nx} = 0$ by taking the linear combination $\mathbf{x}'(\lambda \mathbf{M} + \mu \mathbf{N})\mathbf{x} = 0$. We illustrate this for the circle $x^2 + y^2 - 5 = 0$ (yellow) and hyperbola $x^2 - y^2 - 3 = 0$ (blue). The four common points are the base points of the pencil, and for any other point there is only one conic of the pencil that passes through it.

Pencil of conics through intersection points of circle x^2 + y^2 - 5 = 0 and hyperbola x^2 - y^2 - 3 = 0

Note that in $\mathbb{R}^2$ conics can intersect in 0, 2 or 4 (possibly coincident) points. If two given circles intersect each other at 2 points, the procedure above for the pencil of circles generates all circles passing through their intersection points, including the degenerate case of the radical axis. If the two points are coincident, the generated circles and radical axis will all be mutually tangential there. The procedure for the pencil of conics will similarly find all conics passing through the intersection points of two given conics, but only if they intersect in 4 (possibly coincident) points. Even if the given conics intersect only in 2 points, the pencil gives a method to find the intersection points of two conics: find (one of) the degenerate conics in the pencil by obtaining a solution to $\det (\lambda \mathbf{M} + \mu \mathbf{N}) = 0$. This involves solving a cubic equation in $\mu/\lambda$ (or just $\lambda$ if you use the A-level textbook's approach and solve $\det (\mathbf{M} + \lambda \mathbf{N}) = 0$ instead). Find the equations of the lines that make up your degenerate conic: if you have a degenerate parabola or hyperbola, you'll need to factorise. Now find the intersection of your lines with one of your given conics. In the example illustrated above, the degenerate cases occur when $\lambda:\mu = -1:1$ (giving the degenerate parabola $2-2y^2-0$ which factorises as $(y+1)(y-1)=0$ i.e. the parallel lines $y=\pm 1$), when $\lambda:\mu = -3:5$ (giving the degenerate hyperbola $2x^2 - 8y^2 = 0$ which factorises as $(x+2y)(x-2y)=0$ i.e. the intersecting lines $y=\pm \frac{1}{2}x$) and when $\lambda:\mu=1:1$ (giving the degenerate parabola $2x^2-8=0$ which factorises as $(x+2)(x-2)=0$ i.e. the parallel lines $x=\pm 2$). Clearly the intersection points are $(2,1)$, $(2,-1)$, $(-2,1)$ and $(-2,-1)$.

Now we introduce the notation of Ferdinand Joachimsthal. Define for points $P(x_i, y_i)$ and $Q(x_j, y_j)$ a quantity $s_{ij} = \mathbf{x}_i' \mathbf{Mx}_j$. This is equivalent to taking the standard equation of the conic and transforming the quadratic terms as $x^2 \to x_i x_j$, $y^2 \to y_i y_j$ and $2xy \to x_i y_j + x_j y_i$, and the linear terms as $2x \to x_i + x_j$ and $2y \to y_i + y_j$:

$$s_{ij} = Ax_i x_j + B(x_i y_j + x_j y_i) + C y_i y_j + F(x_i + x_j) + G(y_i + y_j) + H$$

Note the symmetry $s_{ij} = s_{ji}$, and that point $P(x_i,y_i)$ lies on the conic if and only if $s_{ii}=0$. The textbook extract above shows that in the special case of a point outside a circle, and with the $x^2$ and $y^2$ coefficients standardised to one, then $s_{ii}$ is the squared length of the tangent from that point to the circle so naturally becomes zero if we move that point onto the circle. In fact $s_{ii}$ is just the power of a point with respect to the circle, a definition that also extends to the interior of the circle and ties together the intersecting chords theorem, intersecting secants theorem and tangent-secant theorem.

The equation of a general conic can be written as $s=0$ where $s = \mathbf{x'Mx}$ is like $s_{ii}$ but with the specific point $P(x_i,y_i)$ replaced by the general point $(x,y)$, so $s$ is the left-hand sides of equations $(1)$ and $(2)$.

Keeping $P(x_i, y_i)$ as a specific point, but replacing $x_j$ and $y_j$ associated with point $Q$ by a completely general $x$ and $y$, we can also define $s_i = \mathbf{x}_i' \mathbf{Mx}$, i.e.

$$ s_{i} = Ax_i x + B(x_i y + x y_i) + C y_i y + F(x_i + x) + G(y_i + y) + H$$

Then if point $P(x_i, y_i)$ lies on the conic, the equation of the tangent to the conic at $P$ is $s_i = 0$, i.e.

$$ Ax_i x + B(x_i y + x y_i) + C y_i y + F(x_i + x) + G(y_i + y) + H=0$$

The method for finding the tangent to a circle quoted above was just a special case of this.

The condition for the line $PQ$ to be a tangent to the conic is $s_{ij}^2 = s_{ii}s_{jj}$. We can use this condition to find the equations of the tangents to a conic passing through a given point $P(x_i, y_i)$: we replace $Q(x_j,y_j)$ by the general point $(x,y)$ and rewrite the tangency condition as $s_i^2 - s_{ii}s = 0$. The left-hand side is a quadratic in $x$ and $y$. Factorise it into two linear factors: setting each factor to zero gives the equation of a tangent passing through $P(x_i,y_i)$.

If $P(x_i,y_i)$ lies on the tangent to the conic at $Q(x_j,y_j)$, then $$s_{ii} = A(x_i - x_j)^2 + 2B(x_i - x_j)(y_i - y_j) + C(y_i - y_j)^2$$

This generalises the tangent-length formula for a circle quoted above: put $A=C=1$, $B=0$, and the right-hand side is the square of the length $PQ$. The right-hand side is in general equal to $s_{ii} - 2s_{ij} + s_{jj}$: the algebraic proof boils down to using the symmetry of $\mathbf{M}$ and factorising $\mathbf{x}_i' \mathbf{Mx}_i - \mathbf{x}_i' \mathbf{Mx}_j - \mathbf{x}_j' \mathbf{Mx}_i - \mathbf{x}_j' \mathbf{Mx}_j$ as $(\mathbf{x}_i - \mathbf{x}_j)' \mathbf{M} (\mathbf{x}_i - \mathbf{x}_j)$, where $(\mathbf{x}_i - \mathbf{x}_j)'$ is $\begin{bmatrix}x_i - x_j & y_i - y_j & 0 \end{bmatrix}$. Since the third component is zero, only the coefficients $A$, $B$ and $C$ from the top left two-by-two submatrix of $\mathbf{M}$ appear in our final equation. We have $s_{jj} = 0$ since $Q$ lies on the conic, and $s_{ij} = 0$ by the tangency condition, so $s_{ii} - 2s_{ij} + s_{jj}$ reduces to $s_{ii}$ when $PQ$ is a tangent to the conic at $Q$ .

Joachimsthal's notation gives a powerful way to explore poles and polars with respect to a conic. For a given point $P(x_i, y_i)$, there is an associated polar line with respect to the conic, whose equation is simply $s_i = 0$, and we say $P$ is the pole of that line (again, with respect to the given conic). Stothers treats this both algebraically and geometrically. We saw above the special case that when $P$ lies on the conic, its polar line with respect to the conic is the tangent to the conic at $P$. Vice versa, if a line is tangential to the conic then its pole with respect to that conic is the point of tangency. From the symmetry of $s_{ij}=s_{ji}$, it is trivial to prove La Hire's theorem that if $P$ lies on the polar of $Q$, then $Q$ lies on the polar of $P$ (with respect to the same conic). So as a practical example, if a line intersects the conic at two points $P$ and $Q$, and the tangents to the conic at $P$ and $Q$ intersect at $R$, then $R$ is the pole of the original line $PQ$ with respect to the conic. This follows since $R$ lay on both the polar of $P$ and polar of $Q$ so, by La Hire's theorem, $P$ and $Q$ both lie on the polar of $R$, and these two distinct points suffice to define the polar line. Vice versa, given a point through which we can find two tangents to the conic, its polar is the line through the two points of tangency.

There are many other uses for this notation, including finding the common tangents to two conics and determining the minor radius of an ellipse from its major axis and a tangent line, but I'll leave it there. An excellent book for further reading is Geometry by David A. Brannan, Matthew F. Esplen and Jeremy J. Gray, Cambridge University Press (1st edition 2002, 2nd edition 2012).


Before you start manipulating the general form of a conic equation you should be able to recognize whether it is a circle, ellipse, parabola or hyperbola.

In standard form, the two coefficients to examine are A and C.

For circles, $ B^2 - 4AC < 0$, $B=0$ and $A=C$

For ellipses, $ B^2 - 4AC < 0 $ and either $B \neq 0$ or $A \neq C$

For parabolas, $B^2 - 4AC = 0$

For hyperbolas, $ B^2 - 4AC > 0$

  • 1
    $\begingroup$ This is the direction my mind went. I'd also add the criteria for lines. $\endgroup$ Mar 13 at 1:44
  • 2
    $\begingroup$ I don't see how this answers the question you asked, which I thought was 'why do we bother with the general form?'. $\endgroup$
    – Sue VanHattum
    Mar 13 at 3:39
  • 1
    $\begingroup$ Nitpicking: the criteria above applies to nondegenerate conic sections. It is not always easy to determine if the graph of a quadratic equation is degenerate or not, without using rotations of axes. $\endgroup$
    – Taladris
    Mar 13 at 4:13
  • $\begingroup$ @Taladris : Is there any chance you can be convinced to say "This criterion is..." and "These criteria are...", thus "The criterion above applies to..."? $\endgroup$ Mar 15 at 4:45
  • 2
    $\begingroup$ @SueVanHattum "Identify which of the following equations describes (a) an ellipse, (b) a parabola, or (c) a hyperbola" is one of the most common questions you see in an introductory course about conics. This answer explains why writing the equations in standard form allows us to answer this fundamental question easily (at least for nondegenerate conics) so I think that's motivation enough $\endgroup$
    – Silverfish
    Mar 15 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.