Conic Section Equation Calculator – Classify Ellipse, Parabola, Hyperbola
A conic section is a curve described by a quadratic equation in \(x\) and \(y\).
The most general second-degree equation is
\[
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0.
\]
The coefficients \(A,B,C\) control the “quadratic part,” while \(D,E\) translate/tilt the curve and \(F\) shifts it.
1) Classification with the discriminant
The conic type is determined by the discriminant
\[
\Delta = B^2 - 4AC.
\]
This splits the possibilities:
-
\(\Delta < 0\): Ellipse (includes circles as a special case).
-
\(\Delta = 0\): Parabola.
-
\(\Delta > 0\): Hyperbola.
2) Degenerate cases
Sometimes the equation does not produce a smooth conic (it may become a pair of lines, a point, or no real curve).
A common test uses the determinant of the augmented quadratic-form matrix
\[
Q=
\begin{pmatrix}
A & \tfrac{B}{2} & \tfrac{D}{2}\\
\tfrac{B}{2} & C & \tfrac{E}{2}\\
\tfrac{D}{2} & \tfrac{E}{2} & F
\end{pmatrix}.
\]
If \(\det(Q)=0\), the conic is typically degenerate.
Note: If \(A=B=C=0\), the equation is not quadratic at all—it is linear (a line) or inconsistent.
3) Advanced: removing the \(Bxy\) term (rotation)
When \(B\neq 0\), the conic may be rotated relative to the coordinate axes.
A rotation by angle \(\varphi\) can eliminate the cross term \(Bxy\) (diagonalize the quadratic part):
\[
\tan(2\varphi)=\frac{B}{A-C}
\quad\Longrightarrow\quad
\varphi=\frac12\arctan\!\left(\frac{B}{A-C}\right).
\]
In rotated coordinates \((u,v)\), the quadratic part becomes
\[
\lambda_1 u^2 + \lambda_2 v^2,
\]
where \(\lambda_1,\lambda_2\) are the eigenvalues of the matrix
\(\begin{pmatrix}A & B/2\\ B/2 & C\end{pmatrix}\).
4) Advanced: center and a canonical-form hint
For most ellipses and hyperbolas, the conic has a center.
It can be found by solving:
\[
\frac{\partial}{\partial x}(Ax^2+Bxy+Cy^2+Dx+Ey+F)=0,\quad
\frac{\partial}{\partial y}(Ax^2+Bxy+Cy^2+Dx+Ey+F)=0,
\]
which gives the linear system
\[
2Ax + By = -D,\qquad
Bx + 2Cy = -E.
\]
After translating to the center and rotating to remove \(Bxy\), the equation often takes a clean canonical shape
such as
\[
\frac{u^2}{a^2}+\frac{v^2}{b^2}=1 \quad(\text{ellipse}),
\qquad
\frac{u^2}{a^2}-\frac{v^2}{b^2}=1 \quad(\text{hyperbola}).
\]
For parabolas, a “center” does not exist. Instead, after rotation, completing the square can reveal a vertex form like
\((v-v_0)^2 = 4p(u-u_0)\).
5) Example: \(x^2 + 4y^2 = 16\)
Write it in general form:
\[
x^2 + 4y^2 - 16 = 0
\]
so \(A=1\), \(B=0\), \(C=4\), \(D=0\), \(E=0\), \(F=-16\).
The discriminant is
\[
\Delta = B^2 - 4AC = 0 - 4(1)(4) = -16 < 0,
\]
therefore it is an ellipse. Dividing by 16 gives:
\[
\frac{x^2}{16}+\frac{y^2}{4}=1.
\]
Graph note
The plot uses square units (equal scaling in x and y), so ellipses and hyperbolas are not visually distorted.
Tick labels are drawn near the axes so they remain readable during pan/zoom.
The conic type is also printed directly on the graph.
