11.4 General conics, coordinate rotation, and classification

An optional extension beyond the standard forms

This section is optional. It brings the standard parabola, ellipse, and hyperbola equations into one calculation, including equations whose real solution set is a point, several lines, or empty. The aim is to recognize geometry without assuming that the coordinate axes already follow the axes of the curve.

A focus-directrix condition explains why quadratic equations arise. For a focus $(p,q)$, a genuine directrix $ux+vy+w=0$, and eccentricity $e\gt0$, where $u^2+v^2\gt0$ and the focus is off the directrix, the condition is

$$\sqrt{(x-p)^2+(y-q)^2} =e\frac{|ux+vy+w|}{\sqrt{u^2+v^2}}.$$

Both sides are nonnegative, so squaring is reversible. Multiplying by $u^2+v^2$ produces a quadratic equation. The converse question needs more care: an arbitrary quadratic equation need not describe a nondegenerate focus-directrix conic. Its lower-degree terms can change the real locus without changing its quadratic part.

Definition

A real quadratic locus

For real coefficients, write

$$f(x,y)=Ax^2+Bxy+Cy^2+Dx+Ey+F.$$

Its real locus is the set of points satisfying $f(x,y)=0$. The quadratic part is $Ax^2+Bxy+Cy^2$, and its discriminant is $\Delta=B^2-4AC$. We also allow $A=B=C=0$ when discussing the boundary cases, although the equation then has degree at most one.

Multiplying the entire equation by a nonzero constant preserves its locus. Rotating and translating coordinates also preserve distances and angles; they change the description of a fixed set, not its geometric type. A rotation removes the mixed term, and a translation then exposes which constant and linear terms still matter.

Rotating the coordinate axes

Rotate the axes counterclockwise by $\theta$. Put $c=\cos\theta$ and $s=\sin\theta$, and denote the new coordinates by $(X,Y)$. The new unit coordinate vectors, expressed in the old coordinates, are $(c,s)$ and $(-s,c)$. Thus the same point has coordinates related by

$$x=cX-sY,\qquad y=sX+cY; \qquad X=cx+sy,\qquad Y=-sx+cy.$$

The inverse formulas follow by substitution using $c^2+s^2=1$. Keeping track of this direction prevents a common sign error: rotating the axes and actively rotating a point are different descriptions.

Theorem

Removal of the mixed term and rotation invariants

After this substitution, the equation has coefficients

$$\begin{aligned} A'&=Ac^2+Bcs+Cs^2, & C'&=As^2-Bcs+Cc^2,\\ B'&=B\cos2\theta-(A-C)\sin2\theta,\\ D'&=Dc+Es, & E'&=-Ds+Ec,\qquad F'=F. \end{aligned}$$

Some real angle always makes $B'=0$. Moreover,

$$A'+C'=A+C,\qquad B'^2-4A'C'=B^2-4AC.$$

Proof: Choose an angle only after identifying the coefficient to cancel

The goal is to eliminate $XY$, not the linear terms. Expanding $Ax^2$, $Bxy$, and $Cy^2$ gives the combined mixed coefficient $-2Acs+B(c^2-s^2)+2Ccs$, which is the stated $B'$. If $B=0$, choose $\theta=0$. If $B\ne0$ and $A=C$, choose $\theta=\pi/4$. Otherwise choose an angle satisfying $\tan2\theta=B/(A-C)$, with $2\theta$ between $-\pi/2$ and $\pi/2$. Its cosine is nonzero, so multiplication by that cosine verifies $B'=0$. These cases avoid dividing by zero when $A=C$.

For the trace identity, add the formulas for $A'$ and $C'$; the mixed contributions cancel. For the discriminant, subtraction gives $A'-C'=(A-C)\cos2\theta+B\sin2\theta$. Squaring this and $B'$ and adding cancels the cross products. Therefore

$$B'^2+(A'-C')^2=B^2+(A-C)^2.$$

Subtract the already invariant square $(A'+C')^2=(A+C)^2$ to obtain the discriminant identity. The argument proves preservation for every rotation, including the angle chosen to remove the mixed term.

A subsequent translation does not change any quadratic coefficient: expanding a translated square only adds linear and constant terms. Thus $\Delta$ also survives translation. These invariants belong to the chosen polynomial equation; multiplying that equation by a scalar $r$ multiplies its trace by $r$ and its discriminant by $r^2$. In particular, the discriminant's sign remains a property of the same locus description under such a rescaling.

θ = 0°

$5X^2 + 6XY + 5Y^2 = 8$

The fixed ellipse is $5x^2+6xy+5y^2=8$. The dashed axes are the new $X,Y$ directions. Use $x=X\cos\theta-Y\sin\theta$ and $y=X\sin\theta+Y\cos\theta$. At $45^\circ$ the mixed coefficient vanishes and the same ellipse has equation $8X^2+2Y^2=8$. This changes coordinates, not the locus.

Complete the squares and retain all cases

After rotation, rename the coefficients to write

$$\alpha X^2+\gamma Y^2+dX+eY+f=0.$$

Here the lower-case $e$ is a linear coefficient, not the eccentricity used in the motivating definition. The rank of the quadratic part means the number of nonzero coefficients among $\alpha,\gamma$ after removal of the mixed term. There are only three possibilities. We can classify them using ordinary square completion, without a general matrix theory.

Theorem

Classification of the real locus after rotation

If the quadratic part has rank two, put $h=-d/(2\alpha)$, $k=-e/(2\gamma)$ and $m=d^2/(4\alpha)+e^2/(4\gamma)-f$. Translation to $U=X-h$, $V=Y-k$ gives

$$\alpha U^2+\gamma V^2=m.$$

If $\alpha,\gamma$ have the same sign, multiply by $-1$ if necessary so both are positive, also changing $m$'s sign. The locus is an ellipse for $m\gt0$, one point for $m=0$, and empty for $m\lt0$. If the coefficients have opposite signs, the locus is a hyperbola for $m\ne0$, and two intersecting lines for $m=0$.

If the rank is one, interchange the coordinate roles if needed so that $\alpha\ne0$ and $\gamma=0$. Completing just the available square gives

$$\alpha U^2+eY+g=0,\qquad U=X+\frac{d}{2\alpha},\qquad g=f-\frac{d^2}{4\alpha}.$$

For $e\ne0$ this is a parabola. For $e=0$, set $r=-g/\alpha$; there are two parallel lines if $r\gt0$, one double line if $r=0$, and no real points if $r\lt0$.

If the rank is zero, the equation is $dX+eY+f=0$. It is a line when $(d,e)\ne(0,0)$, the whole plane when $d=e=f=0$, and empty otherwise.

To justify the definite case, a sum of positive multiples of squares is nonnegative and vanishes only when both variables vanish. For positive $m$, division gives semiaxes $\sqrt{m/\alpha}$ and $\sqrt{m/\gamma}$. In the indefinite case write $\alpha U^2-\beta V^2=m$ with $\alpha,\beta\gt0$, after an interchange or overall sign change if needed. A nonzero right side can be normalized to either standard hyperbola orientation. A zero right side factors as $(\sqrt\alpha U-\sqrt\beta V)(\sqrt\alpha U+\sqrt\beta V)=0$, giving two distinct intersecting lines.

For rank one with $e\ne0$, set $V=Y+g/e$. Then $\alpha U^2=-eV$, which has a nonzero parabola coefficient. Its opening is determined by the sign of $-e/\alpha$. The term $eY$ cannot be absorbed into a square in $Y$ because there is no $Y^2$ term. If $e=0$, every permitted value of $U$ leaves $Y$ free, so each real root gives an entire line. A double line is one geometric line with a repeated factor in the polynomial. Finally, the rank-zero cases follow by solving a genuine linear equation or checking a constant equation.

Counterexample mode

A positive discriminant does not exclude intersecting lines

The equation $x^2-y^2=0$ has $\Delta=4\gt0$, but factors as $(x-y)(x+y)=0$. Its locus is the union of $y=x$ and $y=-x$, not a nondegenerate hyperbola. Changing only the constant gives $x^2-y^2=1$, which is a hyperbola with the same discriminant.

The repaired statement is that a positive discriminant gives an indefinite quadratic part. Square completion must still decide whether the remaining constant is zero. Similarly, a negative discriminant allows an ellipse, a point, or the empty set; a zero discriminant requires examination of the surviving linear terms and the rank.

A circle requires $A=C\ne0$ and $B=0$. Indeed, the identity used in the proof shows that equal diagonal coefficients and a zero mixed coefficient after rotation force $B^2+(A-C)^2=0$ before rotation as well. In that case dividing by $A$ and completing squares gives center $(-D/(2A),-E/(2A))$ and

$$R^2=\frac{D^2+E^2}{4A^2}-\frac FA.$$

Only $R^2\gt0$ gives a genuine circle. Zero gives its center alone, and a negative value gives no real points. Equal quadratic coefficients alone do not establish a positive radius.

Worked classifications

Worked example

Example 1: a rotated and translated ellipse

Classify and locate the axes of

$$5x^2-4xy+8y^2+4\sqrt5x-16\sqrt5y+4=0.$$

Here $\Delta=-144$, so the quadratic part is definite. Take $c=2/\sqrt5$, $s=1/\sqrt5$, satisfying $\tan2\theta=4/3$. Substituting $x=(2X-Y)/\sqrt5$, $y=(X+2Y)/\sqrt5$ gives

$$4X^2+9Y^2-8X-36Y+4=0 \quad\Longleftrightarrow\quad 4(X-1)^2+9(Y-2)^2=36.$$

The positive right side now confirms a real ellipse. Its center in rotated coordinates is $(1,2)$, hence in original coordinates it is $(0,\sqrt5)$. Its semiaxes are $3$ along $(2,1)/\sqrt5$ and $2$ along $(-1,2)/\sqrt5$. A sketch follows by marking the center and adding and subtracting these scaled unit directions to locate the four axis endpoints. The discriminant alone would not supply the center or lengths.

Worked example

Example 2: the linear term distinguishes a parabola from parallel lines

Consider $x^2+2xy+y^2-2x+2y=0$. Although $\Delta=0$, do not discard the linear part. With $\theta=\pi/4$, we have $x+y=\sqrt2X$ and $-2x+2y=2\sqrt2Y$, so the equation becomes $2X^2+2\sqrt2Y=0$, a parabola opening in the negative $Y$ direction.

Compare $x^2+2xy+y^2-2=0$. The same rotation gives $X^2=1$, two parallel lines $x+y=\pm\sqrt2$. Replacing $-2$ by $0$ gives the double line $x+y=0$; replacing it by $2$ gives the empty set. All four equations have the same quadratic part, so their different loci must be determined from the remaining terms.

Worked example

Example 3: one quadratic part and three definite outcomes

The equations $x^2+y^2-2x+4y+c=0$ all have $\Delta=-4$. Completing squares gives $(x-1)^2+(y+2)^2=5-c$. For $c=1$ the locus is a circle of radius $2$ centered at $(1,-2)$. For $c=5$ it is just $(1,-2)$, and for $c=6$ it is empty. The same center calculation is useful in each case, but a radius cannot be assigned by taking a real square root when the right side is negative.

The quadratic matrix viewpoint

Concept lensAlgebraic

A symmetric matrix records the same quadratic expression

Define column vectors $z=(x,y)^T$, $Z=(X,Y)^T$ and

$$Q=\begin{pmatrix}A&B/2\\B/2&C\end{pmatrix},\qquad O=\begin{pmatrix}c&-s\\s&c\end{pmatrix}.$$

Then $z^TQz=Ax^2+Bxy+Cy^2$: the two off-diagonal products each contribute half of the mixed term. The coordinate change is $z=OZ$. Since $O^TO=OO^T=I$, substitution yields

$$z^TQz=Z^T(O^TQO)Z,\qquad Q'=O^TQO, \qquad Q=OQ'O^T.$$

The last identity follows by multiplying on the left by $O$ and on the right by $O^T$. Matrix multiplication is ordered; these two formulas are inverse transformations, not interchangeable patterns. Choosing the angle that makes $B'=0$ makes $Q'$ diagonal. This is the matrix description of the rotation already proved by trigonometry.

The trace is the sum of the diagonal entries, $A+C$, and the determinant is $AC-B^2/4=-\Delta/4$. Because $\det O=1$, taking determinants in $Q'=O^TQO$ preserves the determinant; the trace preservation is already proved by the coefficient calculation. Thus the matrix packages the same two invariants. The linear term transforms separately to $(D,E)OZ$, so the matrix of the quadratic part alone cannot decide all real-locus cases. No general diagonalization theorem is needed here.

Tangents require a regular point

Theorem

The tangent formula at a regular point of a quadratic locus

Let $P=(x_0,y_0)$ satisfy $f(P)=0$, and suppose

$$\nabla f(P)=(2Ax_0+By_0+D,\ Bx_0+2Cy_0+E)\ne(0,0).$$

Then the tangent line is $\nabla f(P)\cdot(x-x_0,y-y_0)=0$, or equivalently

$$Ax_0x+\frac B2(x_0y+y_0x)+Cy_0y +\frac D2(x+x_0)+\frac E2(y+y_0)+F=0.$$

Differentiate $f(x(t),y(t))=0$ along a regular local parametrization. The chain rule gives $\nabla f(P)\cdot(x'(t),y'(t))=0$ at the contact. A nonzero gradient supplies a genuine normal direction. If its second component is nonzero, implicit differentiation gives a local graph $y=y(x)$; if only its first component is nonzero, use $x=x(y)$ instead. Thus a vertical tangent does not require division by zero.

To obtain the symmetric formula, expand the point-normal equation, use $f(P)=0$ to replace its constant term, and divide by two. Both hypotheses matter. A point outside the locus has no tangent there, and at a singular point the gradient equation is $0=0$. For example, the origin on $x^2-y^2=0$ lies on two crossing branches; the displayed formula cannot choose one tangent direction for them.

Quick checks

Checkpoint

Check 1: what if the diagonal coefficients agree?

For $A=C$, how can the mixed term be removed without dividing by $A-C$?

Solution · Answer to Check 1

If $B=0$, use $\theta=0$. Otherwise use $\theta=\pi/4$; then $B'=B\cos(\pi/2)=0$. The choice does not require a tangent quotient.

Checkpoint

Check 2: is a zero discriminant enough?

Classify $x^2-1=0$ and $y-1=0$. Why can both have $\Delta=0$?

Solution · Answer to Check 2

The first is two parallel lines $x=\pm1$ and has rank one. The second is the line $y=1$ and has rank zero. A zero discriminant records a singular quadratic part, without specifying its rank or linear terms.

Exercises

1. Rotate $xy=1$ by $\theta=\pi/4$ and identify the resulting standard form. What changes if the equation is $xy=0$?
2. Classify $x^2+4y^2-2x+16y+17=0$ and $x^2+4y^2-2x+16y+18=0$.
3. Find the tangent to $x^2+4y^2=4$ at $(0,1)$ and at $(2,0)$. Explain why the second tangent is still covered by the theorem.
4. Classify the rank-zero equations $2x-y+3=0$, $0=0$, and $1=0$. State which of these has a nonzero gradient.

Guided solutions

Solution · Solution 1

The substitution gives $xy=(X^2-Y^2)/2$. Thus $xy=1$ becomes $X^2/2-Y^2/2=1$, a hyperbola with semiaxes $\sqrt2$. For $xy=0$, the result is $X^2-Y^2=0$, two intersecting lines. In the original coordinates these are the coordinate axes.

Solution · Solution 2

The left sides become $(x-1)^2+4(y+2)^2$ and $(x-1)^2+4(y+2)^2+1$, respectively. Hence the first locus is the single point $(1,-2)$ and the second is empty. Both have $\Delta=-16$, which identifies a definite quadratic part but does not assert a nondegenerate ellipse.

Solution · Solution 3

The gradient is $(2x,8y)$, giving $(0,8)$ at $(0,1)$ and $(4,0)$ at $(2,0)$. The tangent equations are $y=1$ and $x=2$. Both normals are nonzero. The point-normal formula covers the vertical line directly, without representing it as a graph of $y$ against $x$.

Solution · Solution 4

The loci are a line, the whole plane, and the empty set, respectively. Only the first polynomial has a nonzero gradient, namely $(2,-1)$. The identically zero polynomial imposes no restriction at all; the constant polynomial $1$ has no zero at which to consider a tangent.