Why Eccentricity Classifies Conics
A polar equation can describe very different conic curves even when its overall form stays the same. The key distinguishing number is the eccentricity, written \(e\). Comparing \(e\) with \(1\) classifies the conic as an ellipse, a parabola, or a hyperbola. The circle appears as the special case \(e=0\).
In this tutorial, we use the standard polar form for a conic with its focus at the pole and its axis aligned with the polar axis:
Here \(\ell>0\) is the semi-latus rectum, and \(e\geq 0\) is the eccentricity. A conic oriented in another direction may use \(\sin\theta\), or a shifted angle, instead of \(\cos\theta\). That change in orientation does not change the eccentricity or the classification.
The Classification Rule
- If \(0\leq e<1\), the conic is an ellipse. When \(e=0\), it is a circle.
- If \(e=1\), the conic is a parabola.
- If \(e>1\), the conic is a hyperbola.
The comparison is about the value of \(e\), not the size of \(\ell\). Changing \(\ell\) changes the scale of the conic, but does not change whether it is an ellipse, parabola, or hyperbola.
| Eccentricity | Classification | Special note |
|---|---|---|
| \(e=0\) | Circle | A special ellipse centered at the pole in this form |
| \(0<e<1\) | Ellipse | Not a circle |
| \(e=1\) | Parabola | The boundary between the two other classifications |
| \(e>1\) | Hyperbola | The standard positive-radius equation describes a branch |
Why the Rule Works
A short conversion to Cartesian coordinates explains the thresholds. Write \(x=r\cos\theta\) and \(r=\sqrt{x^2+y^2}\). Starting with the standard polar equation and multiplying by its denominator gives
Rearranging and squaring yields
so the Cartesian equation is
The important feature is the coefficient \(1-e^2\) of \(x^2\). When it is positive, the quadratic terms have the same sign; when it is zero, the \(x^2\) term disappears; and when it is negative, the quadratic terms have opposite signs. Completing the square makes each case explicit.
Since \(1-e^2>0\), completing the square gives \((1-e^2)\left(x+\frac{\ell e}{1-e^2}\right)^2+y^2=\frac{\ell^2}{1-e^2}\). This is an ellipse. If \(e=0\), it reduces to \(x^2+y^2=\ell^2\), a circle.
The \(x^2\) term vanishes, leaving \(y^2+2\ell x-\ell^2=0\), or \(y^2=-2\ell\left(x-\frac{\ell}{2}\right)\). This is a parabola.
Now \(e^2-1>0\), and completing the square gives \((e^2-1)\left(x-\frac{\ell e}{e^2-1}\right)^2-y^2=\frac{\ell^2}{e^2-1}\). This is the standard form of a hyperbola.
Squaring can introduce points that do not satisfy the original equation: the original relation also requires \(\ell-ex\geq0\), since \(\sqrt{x^2+y^2}=\ell-ex\). For \(e>1\), this condition selects the branch represented by the standard positive-radius polar equation. It does not alter the hyperbola classification.
How to Read Eccentricity from an Equation
Before classifying, put the equation into standard form. For example, if the equation is \(r=\frac{A}{B+C\cos\theta}\), with \(A>0\) and \(B>0\), factor \(B\) from the denominator:
Thus \(\ell=A/B\), and the eccentricity is \(e=|C/B|\). The absolute value is important: a negative coefficient changes the orientation of the conic, not the nonnegative value of its eccentricity. A sine term works the same way. More generally, if the constant term in the denominator is not \(1\), divide it out before identifying \(e\).
Worked Example: Recognizing a Circle
Classify \(r=\frac{8}{2+0\cos\theta}\).
First simplify the denominator and normalize:
Therefore \(\ell=4\) and \(e=0\). Since \(e=0<1\), this is the circle case. The polar equation is \(r=4\), a circle of radius \(4\) centered at the pole.
Check: substituting \(\ell=4\) and \(e=0\) into the standard form gives \(r=\frac{4}{1+0\cos\theta}=4\), exactly the original simplified equation.
Worked Example: Classifying an Ellipse
Classify \(r=\frac{10}{2+\cos\theta}\).
Factor \(2\) from the denominator:
Thus \(\ell=5\) and \(e=\frac12\). Since \(0<\frac12<1\), the conic is an ellipse, not a circle.
Check: inserting \(\ell=5\) and \(e=\frac12\) into the standard form gives \(r=\frac{5}{1+\frac12\cos\theta}\). Multiplying numerator and denominator by \(2\) recovers \(r=\frac{10}{2+\cos\theta}\).
Worked Example: Identifying a Parabola
Classify \(r=\frac{12}{4+4\sin\theta}\).
Factor \(4\) from the denominator and simplify:
This is the sine version of standard form, with \(\ell=3\) and \(e=1\). Therefore the conic is a parabola.
Check: \(\frac{\ell}{1+e\sin\theta}=\frac{3}{1+\sin\theta}\), which agrees with the normalized equation. At angles where \(1+\sin\theta=0\), the radius is undefined; this is consistent with the curve extending without bound in that direction, not evidence for a different eccentricity.
Worked Example: Classifying a Hyperbola
Classify \(r=\frac{15}{3-6\cos\theta}\).
Factor \(3\) from the denominator:
The coefficient of \(\cos\theta\) is \(-2\), so the eccentricity is its absolute value, \(e=2\), and \(\ell=5\). Since \(2>1\), the conic is a hyperbola.
Check: the standard form with \(\ell=5\) and \(e=2\) is \(r=\frac{5}{1+2\cos\theta}\). Reversing the direction of the polar axis changes \(\cos\theta\) to \(-\cos\theta\), giving the displayed equation. The sign changes orientation; the eccentricity remains \(2\).
A Common Mistake: Reading the Wrong Coefficient
A frequent error is to read the coefficient of the trigonometric term before normalizing the denominator. In \(r=\frac{10}{2+\cos\theta}\), it is incorrect to say \(e=1\) just because the cosine coefficient is \(1\). The constant term is \(2\), so the denominator must first be divided by \(2\): \(r=\frac{5}{1+\frac12\cos\theta}\). The eccentricity is therefore \(\frac12\), and the conic is an ellipse, not a parabola.
Why Classification Is Useful
Classifying a conic before trying to interpret its graph gives a reliable first conclusion from the equation alone. In particular, it prevents confusing a change of scale with a change of curve type: \(\ell\) sets the scale, while \(e\) determines the classification. The equation’s orientation may change when a cosine becomes a sine or its coefficient changes sign, but the comparison with \(1\) remains the same.
When an equation is not already normalized, take the time to factor its denominator and identify the coefficient relative to the constant term. That one step avoids a common misclassification and provides a direct check against the standard form.
Check Your Understanding
For each equation, normalize it if needed, identify its eccentricity, and state its classification.
- Classify \(r=\frac{6}{1+\frac{3}{4}\cos\theta}\). Is it a circle?
- Classify \(r=\frac{8}{2-2\sin\theta}\).
- For \(r=\frac{9}{3+6\cos\theta}\), what is \(e\), and what is the conic?
- Does changing \(r=\frac{\ell}{1+e\cos\theta}\) to a form with \(\sin\theta\) change its eccentricity?
- Why must a denominator such as \(2+3\cos\theta\) be normalized before reading off \(e\)?
