Reading orientation from a polar conic
A polar conic with a focus at the pole is commonly written with a denominator involving either cosine or sine. The trigonometric term tells you which axis contains the conic’s vertices; its sign tells you which side of the pole contains the directrix. Together with the eccentricity, these features give a reliable starting point for a sketch.
Here we use \(d>0\) for the distance from the pole to the directrix and \(e>0\) for the eccentricity. The radius \(r\) is allowed to be negative: a negative polar radius represents a point in the direction opposite to the stated angle. That convention matters especially when locating the vertices of a hyperbola.
Cosine, sine, and the directrix
The link between the equation and the directrix follows from the focus-directrix definition of a conic: the distance from a point to the focus is \(e\) times its perpendicular distance to the directrix. With the focus at the pole, the focus distance is \(r\) when \(r\geq 0\), and the Cartesian coordinates are \(x=r\cos\theta\) and \(y=r\sin\theta\).
For a directrix \(x=d\), consider points on the side \(x<d\). Their perpendicular distance to the directrix is \(d-x=d-r\cos\theta\). The focus-directrix relation becomes \(r=e(d-r\cos\theta)\). Rearranging gives the cosine form below. This argument explains why a plus sign in the denominator goes with a directrix on the positive \(x\)-side.
The derivation uses the stated side of the directrix; for conics that have points on both sides, signed-radius values help represent the full curve. Replacing \(x\) by \(y\), or equivalently rotating the axis, gives the sine form. A minus sign corresponds to placing the directrix at \(x=-d\) or \(y=-d\), respectively.
| Polar form | Axis containing vertices | Directrix |
|---|---|---|
| \(r=\frac{ed}{1+e\cos\theta}\) | \(x\)-axis | \(x=d\) |
| \(r=\frac{ed}{1-e\cos\theta}\) | \(x\)-axis | \(x=-d\) |
| \(r=\frac{ed}{1+e\sin\theta}\) | \(y\)-axis | \(y=d\) |
| \(r=\frac{ed}{1-e\sin\theta}\) | \(y\)-axis | \(y=-d\) |
Vertices from the cosine form
For \(r=\frac{ed}{1+e\cos\theta}\), the axis is horizontal, so the vertices occur at the axial angles \(\theta=0\) and \(\theta=\pi\). Substituting these angles gives the signed radii. To locate the actual points, use \(x=r\cos\theta\), not just the value of \(r\).
For an ellipse, \(0<e<1\), both radii are positive. Its vertices are on opposite sides of the pole: one at positive \(x\), the other at negative \(x\). The vertex at \(\theta=\pi\) is farther from the pole, since \(\frac{ed}{1-e}>\frac{ed}{1+e}\). For a parabola, \(e=1\), the denominator vanishes at \(\theta=\pi\); the curve extends without bound in that direction. For a hyperbola, \(e>1\), the radius at \(\theta=\pi\) is negative. Its point is therefore on the positive \(x\)-axis, at \(x=\frac{ed}{e-1}\). Both vertices are on the positive side of the pole, and the hyperbola’s axis is horizontal.
Worked Example: Sketching a cosine ellipse
Worked Example: Sketching a cosine ellipse
Sketch the main features of \(r=\frac{2}{1+\frac12\cos\theta}\).
Match the equation to \(r=\frac{ed}{1+e\cos\theta}\). The coefficient in the denominator gives \(e=\frac12\), and \(ed=2\). Therefore \(d=4\), since \(2=(\frac12)d\). Because \(0<e<1\), this is an ellipse. The cosine term gives a horizontal axis, and the plus sign places the directrix at \(x=4\).
At \(\theta=0\), the denominator is \(1+\frac12=\frac32\), so \(r=2/(\frac32)=\frac43\). The vertex is \((\frac43,0)\). At \(\theta=\pi\), the denominator is \(1-\frac12=\frac12\), so \(r=2/(\frac12)=4\). Since \((x,y)=(r\cos\theta,r\sin\theta)\), this vertex is \((-4,0)\).
As a check, at \((\frac43,0)\) the focus distance is \(\frac43\) and the distance to \(x=4\) is \(4-\frac43=\frac83\). Their ratio is \((\frac43)/(\frac83)=\frac12=e\). At \((-4,0)\), the focus distance is \(4\) and the directrix distance is \(8\); their ratio is \(4/8=\frac12\). The vertices satisfy the focus-directrix condition. A sketch should show an ellipse with horizontal major axis, vertices at \((-4,0)\) and \((\frac43,0)\), focus at the pole, and directrix \(x=4\).
Worked Example: Orienting a parabola
Worked Example: Orienting a parabola
Find the vertex and sketch the orientation of \(r=\frac{3}{1+\cos\theta}\).
Here \(e=1\) and \(ed=3\), so \(d=3\). The directrix is \(x=3\), and the cosine term makes the axis horizontal. At \(\theta=0\), \(r=3/(1+1)=\frac32\), giving the vertex \((\frac32,0)\).
At \(\theta=\pi\), the denominator is zero, so there is no finite radius in that direction. This is the unbounded direction of the parabola. In particular, the curve extends opposite the directrix: it opens toward negative \(x\). One check comes from the focus-directrix equation. For a point \((x,y)\) on the side \(x\leq 3\), its focus distance is \(\sqrt{x^2+y^2}\), while its distance to the directrix is \(3-x\). With \(e=1\), squaring \(\sqrt{x^2+y^2}=3-x\) gives \(x^2+y^2=(3-x)^2\), hence \(y^2=9-6x\). This has vertex \((\frac32,0)\) and grows as \(x\) decreases, confirming the leftward orientation.
Worked Example: Locating hyperbola vertices
Worked Example: Locating hyperbola vertices
Determine the directrix, vertices, and axis direction for \(r=\frac{2}{1+2\cos\theta}\).
Here \(e=2\) and \(ed=2\), so \(d=1\). Since \(e>1\), the conic is a hyperbola. It has a horizontal axis and directrix \(x=1\).
At \(\theta=0\), the denominator is \(1+2=3\), so \(r=\frac23\), giving the vertex \((\frac23,0)\). At \(\theta=\pi\), the denominator is \(1-2=-1\), so \(r=-2\). The point is \((x,y)=(r\cos\pi,r\sin\pi)=(2,0)\), not \((-2,0)\). Thus the vertices are \((\frac23,0)\) and \((2,0)\). The hyperbola has two branches extending along the horizontal axis.
Check the focus-directrix ratio at the first vertex: its focus distance is \(\frac23\), and its distance from \(x=1\) is \(\frac13\), so the ratio is \(2\). At the second vertex, those distances are \(2\) and \(1\), again giving \(2\). Both points meet the required ratio \(e=2\).
Rotating the axis and reflecting the directrix
The sine forms follow the same procedure, with vertical in place of horizontal. For instance, \(r=\frac{ed}{1+e\sin\theta}\) has directrix \(y=d\), and its axial vertices occur at \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\). At these angles, \(y=r\sin\theta\), so the actual vertex coordinates must account for both the radius and the sign of the sine.
Changing \(+\cos\theta\) to \(-\cos\theta\) reflects the directrix from \(x=d\) to \(x=-d\); changing \(+\sin\theta\) to \(-\sin\theta\) reflects it from \(y=d\) to \(y=-d\). The axis remains the same. For a quick check on an equation with a minus cosine, substitute \(\theta=0\) and \(\theta=\pi\) into the denominator and then convert each signed radius to Cartesian coordinates. Do not infer the vertex side from the radius alone.
A dependable sketching sequence
For a standard polar conic with its focus at the pole, use the following order. It separates the information encoded by the equation and reduces sign errors.
A cosine term indicates the \(x\)-axis; a sine term indicates the \(y\)-axis.
The sign in the denominator gives the side: \(1+e\cos\theta\) means \(x=d\), while \(1-e\cos\theta\) means \(x=-d\). Apply the corresponding rule for sine.
Use \(e<1\), \(e=1\), or \(e>1\) to identify ellipse, parabola, or hyperbola. Substitute the axial angles and convert signed radii to points.
Place the pole, directrix, and vertices first. Show a parabola extending opposite its directrix; for an ellipse or hyperbola, use the vertex positions and eccentricity to guide the shape.
At each calculated vertex, compare focus distance with directrix distance and verify that their ratio is \(e\).
Common mistake: treating signed radius as a coordinate
A frequent error is to set \(x=r\) at \(\theta=\pi\). The Cartesian conversion is \(x=r\cos\theta\), and \(\cos\pi=-1\). For the hyperbola in the worked example, substituting \(\theta=\pi\) gives \(r=-2\); the coordinate is \(x=(-2)(-1)=2\). The incorrect calculation \(x=r=-2\) puts the vertex on the wrong side of the pole and contradicts the directrix-distance check.
Another useful distinction is that “opening direction” is most direct for a parabola. An ellipse is closed, so describe its axis and the positions of its near and far vertices. A hyperbola has two branches; describe its axis and locate both vertices, taking signed radii into account. In every case, the directrix identifies the orientation more reliably than the visual appearance of a partial sketch.
Check Your Understanding
Use the sign, eccentricity, and axial-angle substitutions to answer each question. Convert any negative radius to its actual point.
- For \(r=\frac{6}{1+\frac12\cos\theta}\), identify the conic type and directrix, then find both vertices.
- For \(r=\frac{4}{1+\sin\theta}\), what is the axis and directrix, and where is the finite vertex?
- For \(r=\frac{3}{1+2\cos\theta}\), calculate the radius at \(\theta=\pi\) and give the corresponding Cartesian point.
- Why does a zero denominator at an axial angle indicate an unbounded direction rather than a finite vertex?
- At a proposed vertex, what distance ratio should be checked to confirm the focus-directrix relation?
