Conic Section Converter – General to Standard Form Calculator
A general conic in the plane can be written as
\[
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0.
\]
This tool identifies the conic type and converts the equation into a standard (often “completed square”) form.
1) Identify the conic type
The discriminant
\[
\Delta = B^2 - 4AC
\]
classifies the conic:
- \(\Delta < 0\) ⇒ ellipse (circle is a special case)
- \(\Delta = 0\) ⇒ parabola
- \(\Delta > 0\) ⇒ hyperbola
2) Degenerate cases
Sometimes the “conic” collapses into lines, a single point, or has no real graph. A common degeneracy test uses
\[
Q=
\begin{pmatrix}
A & \tfrac{B}{2} & \tfrac{D}{2}\\
\tfrac{B}{2} & C & \tfrac{E}{2}\\
\tfrac{D}{2} & \tfrac{E}{2} & F
\end{pmatrix}.
\]
If \(\det(Q)=0\), the equation is typically degenerate.
If \(A=B=C=0\), then the equation is not quadratic at all (it is linear), so it is not a conic section.
3) Remove the \(Bxy\) term (rotation)
When \(B\neq 0\), the conic is rotated. We can rotate axes by an angle \(\varphi\) so the cross term disappears:
\[
\tan(2\varphi)=\frac{B}{A-C}
\quad\Longrightarrow\quad
\varphi=\frac12\arctan\!\left(\frac{B}{A-C}\right).
\]
In rotated coordinates \((u,v)\), the quadratic part becomes diagonal:
\[
\lambda_1 u^2 + \lambda_2 v^2
\]
where \(\lambda_1,\lambda_2\) are eigenvalues of
\(\begin{pmatrix}A & B/2\\ B/2 & C\end{pmatrix}\).
4) Translate to the center (ellipses & hyperbolas)
Most ellipses and hyperbolas have a center \((h,k)\). It is found by solving the system obtained from the partial derivatives:
\[
2Ax + By = -D,\qquad
Bx + 2Cy = -E.
\]
After translating \(X=x-h,\;Y=y-k\), the linear terms vanish and the equation simplifies to a centered quadratic plus a constant.
Then rotating gives a clean standard form:
-
Ellipse (possibly rotated):
\[
\frac{u^2}{a^2}+\frac{v^2}{b^2}=1
\]
-
Hyperbola (possibly rotated):
\[
\frac{u^2}{a^2}-\frac{v^2}{b^2}=1
\quad\text{or}\quad
\frac{v^2}{a^2}-\frac{u^2}{b^2}=1
\]
5) Parabola: vertex form
Parabolas do not have a center. After rotation (if needed), the equation can be rearranged by completing the square to get a vertex form such as
\[
(v-v_0)^2 = 4p(u-u_0)
\quad\text{or}\quad
(u-u_0)^2 = 4p(v-v_0),
\]
where \((u_0,v_0)\) is the vertex (in rotated coordinates) and \(p\) controls the “width” of the parabola.
6) Worked example
Convert
\[
4x^2 + y^2 - 8x + 2y + 1 = 0.
\]
Group and complete squares:
\[
4(x^2-2x) + (y^2+2y) + 1 = 0
\]
\[
4\big((x-1)^2-1\big) + \big((y+1)^2-1\big) + 1 = 0
\]
\[
4(x-1)^2 + (y+1)^2 = 4
\]
Divide by 4:
\[
\frac{(x-1)^2}{1} + \frac{(y+1)^2}{4} = 1.
\]
This is an ellipse centered at \((1,-1)\).
Graph note
The plot uses square units (equal x/y scaling) so the shape is not distorted, and tick labels are drawn near their axes during pan/zoom.
The computed conic type is printed both in the solution and directly on the graph.
