• Trigonometry > Conics from a Circle

Generating Conics from a Circle and a Line

The conic sections can be described by their rectangular equations and polar equations. In the polar form, they are defined as a locus of distances in a ratio with respect to a point, the focus, and a line, the directrix.

I have found a similar but different approach to defining the conics based on a circle and a tangent line. The beauty of this method is that the ratio of the distances is the same eccentricity as the classic definition. Moreover, with this form, one can still let e = 0 and the conic gives us a circle.

Presented here are the rectangular and polar equations of conics as defined by a circle and its “tangent directrix”.

Deriving the Equation of the Conics

The equations I have given above were based on the circle centered at (0, R). And the line was y = 2R. This was years ago. Today, I have positioned the circle at the origin and the line at y = –R. The image below shows the image of the conic formed in orange.

Let’s recall that we are defining our curve such that the distance from a point on the curve to the closest point on the circle, d₁, and the distance from that point to the line tangent to the circle, d₂, is a constant e. (In the above image, PD = d₂, but we have simply called it d. This makes d₁ or CP equal to ed. Point C is the point on the circle that is closest to point P because it lies on a line through the center of the circle.)

So, our problem is to find all points P(x, y), such that CP/PD = e. We can find this relationship in the polar form first, which is easier. Let R be the radius of the circle. Let r be OP, the distance from the origin to point P. First, note that \(PD = r\sin\theta + R\). Also note that \(CP = r - R\). From what we have defined above, we can say CP = ePD. Substition gives us the following steps:

(i) \(e(r\sin\theta + R) = r - R\)

We simply solve for r and get:

(ii) \(r(\theta) = \frac{R(1+e)}{1-e\sin\theta}\)

Note that the classic conic defined by a focal point and a line (the directrix) has the polar form: \(r(\theta)=\frac{ed}{1-e\sin\theta}\), where d is the distance from the point to the directrix and e is the eccentricity. This form does not allow e to be 0, which would give us a circle. However, our equation above allows all values of e, even negative values. And e as we have defined is still the eccentricity of the conic. This is remarkable.

Notice that when we let e = 0, our equation becomes \(r(\theta) = R\), the polar equation of a circle.

The Rectangular Form

When converting the polar form to the rectangular form, we get the following intermediate form:

(i) \(x^2+(1-e^2)y^2-2Re(1+e)y = \) \(R^2(1+e)^2\)

This is suitable for a circle or parabola, which reduces to \(x^2 + y^2 = R^2\) or \(x^2 - 4Ry = 4R^2\) when e = 0 or 1 respectively.

For an ellipse or hyperbola, we can complete the square (calculations are a bit messy) and get the following:

The above equation works for every conic except the parabola. For a hyperbola when e > 1, the denominator of x² becomes negative, giving us opposite signs of the x and y terms.

The Geogebra activity below shows all of the conics graphed. The ratios are the eccentricities of the respective conics. Move point C on the circle to see how the points are traced on the conic and notice that the ratios remain the same.

Also notice that the hyperbola has 2 points where the ratios are equal. The second point, H₃ is on the second branch. This ratio is a little different. Rather than the close point on the circle, it is the farthest point on the circle, or the closest point plus twice the radius. You may need to zoom in and move the canvas to see the second point.

The activity above shows an ellipse with e = 0.5. The two circles are shown as orange and purple. You can change the eccentricity to make it a hyperbola. For a hyperbola, the distance changes from the closest point on the circle to the farthest point. Therefore, there are two ratios of CP/PD in the activity.

You may notice that, when dragging point C on the circle, when C creates an angle that is parallel to the asymptote of the hyperbola, point H₁ no longer falls on the hyperbola.

a, b, c, e, and R

We have the equation in standard form. For ellipses and hyperbolas, we would like to know the values of a, b, and c. The first two are easy enough. \(a = \sqrt{\frac{1+e}{1-e}}R\) and \(b = \frac{1}{1-e}R\). Let’s calculate c for an ellipse. Since the y-axis is the major axis, we will let a be the denominator of the y term.

(i) \(c^2 =\frac{1}{(1-e)^2}R^2 - \frac{1+e}{1-e}R^2 = \) \(\frac{1}{(1-e)^2}R^2 - \frac{1-e^2}{(1-e)^2}R^2 = \) \(\frac{e^2}{(1-e)^2}R^2\)

Therefore, for an ellipse, \(c = \frac{e}{1-e}R\).

For a hyperbola, we need to add a² and b² instead. However, remember that the x² term is negative. Therefore, we need to take the positive value of its denominator. Let’s do the calculation after taking out the negative value if the denominator:

(i) \(c^2 =\frac{1}{(e-1)^2}R^2 + \frac{e+1}{e-1}R^2 = \) \(\frac{1}{(e-1)^2}R^2 + \frac{e^2-1}{(e-1)^2}R^2 = \) \(\frac{e^2}{(e-1)^2}R^2\)

Therefore, for a hyperbola, \(c = \frac{e}{e-1}R\).

Lastly, given a conic in standard form, we want to determine the radius, R of the circle that would generate this conic. We can utilize the values of a, b, and e to determine our radius. Since our conic’s major axis is the y-axis, its denominator will be our a term. We can use the a value to find e in terms of a and R. Then substitute e into the b value and solve for R.

(i) \(a^2 = \frac{1}{(1-e)^2}R^2\)

(ii) \(e = 1 - \frac{R}{a}\)

Substituting into the b value:

(iii) \(b^2 = \frac{(1+1-\frac{R}{a})R^2}{1-(1-\frac{R}{a})} = \) \(\frac{(2-\frac{R}{a})R^2}{\frac{R}{a}}\)

(iv) \(b^2 = 2aR - R^2\)

Solving for R in the quadratic equation, we get:

(v) \(R=a\pm\sqrt{a^2-b^2} = a \pm c\)

These two are the distances from the vertex (that is tangent to our circle and the tangent directrix) to the two foci of the conic.

One Conic, Two Circles

At first glance, R = a – c is the answer that makes most sense since the center of the circle is one focus and it cannot be bigger than length of two foci. Or can it? It turns out, there are 2 circles that give us a constant ratio of e: one circle with radius a – c and another with radius a + c. This is a good place for a Geogebra activity that shows both circles.

As you can see from the activity above, there are potentially 4 ratios that give the eccentricity for ellipses. The circle with radius a – c is the smaller orange circle. The bigger circle in purple has a radius of a + c. There is the points I_c on the “inner” circle closest to the point on the conic P_c and point I_f on the “inner” circle that is farthest from P_c and this intersects the conic again at point P_f. The same goes for the points on the “outer” circle. Thus, 4 ratios are shown above. You can drag points P_c and Q_c to see how the ratios remain the same. The program is a bit wonky with hyperbolas since the distances are so large. However, not all 4 ratios will be equal in hyperbolas but it seems at least 2 of them are.

You may notice that the center of the both circles, C_i and C_o, are the foci of the ellipse.

One Circle, Two Conics

We showed 2 circles and 1 conic that gives 4 ratios that are equal to e above. And we gave the radius lengths. We can also find two conics for 1 circle instead. One conic that is bigger than the circle and one that is smaller. The equation for the bigger conic was stated above: \(r(\theta) = \frac{R(1+e)}{1-e\sin\theta}\). The smaller conic is: \(r(\theta) = \frac{R(1-e)}{1+e\sin\theta}\).

Proof of the Eccentricity

We must prove the eccentricity of the ellipse and hyperbola match the conventional definition. (For the circle and the parabola, this is trivial.) We know that the eccentricity is equal to c/a. Let’s plug in these values (as the squares) and see what we get. In our conic above, the major axis is the y-axis, so a is the denominator of the y² term and b is the denominator of the x² term. The value c² is a² – b². So, we have:

(i) \(\frac{c^2}{a^2} = \frac{\frac{1}{(1-e)^2}R^2 - \frac{1+e}{1-e}R^2}{\frac{1}{(1-e)^2}R^2}=\) \(\frac{\frac{1}{(1-e)^2} - \frac{(1+e)(1-e)}{(1-e)^2}}{\frac{1}{(1-e)^2}}=\) \(\frac{\frac{1 - 1 + e^2}{(1-e)^2}}{\frac{1}{(1-e)^2}}= e^2\)

(ii) \(\sqrt{\frac{c^2}{a^2}} = \sqrt{e^2}\)

(iii) \(\frac{c}{a} = e\)

This is what we needed to show.

Circling Back

The image below shows two ways to calculate the ratio e: the conventional method and the circle method.

In the image, the blue line is the conventional directrix. The red line is our tangent directrix. The ratio OA to AD is equal to the ratio CA to AE is equal to e. Moreover, lines CE and OD are parallel. Therefore, triangles ACE and AOD are similar.

You can interact with the Geogebra activity below to see the effect of dragging points A and B.

Point B is for negative eccentricities.

Similarities and Differences

There are some similarities and differences between the two methods of defining conics.

Of course, the eccentricity is the same for both methods. The focal point is the same for both methods.

One major difference is that the directrix for the conventional method shifts position. Our tangent directrix remains in place. Moreoever, the tangent directrix is tangent to both the conic at its vertex and the circle.

In both polar equations of the conics, the angle that gives us the two ways to calculate e is the same. The conventional equation is \(r(\theta) = \frac{ed}{1-e\sin\theta}\). Our new equation is \(r(\theta) = \frac{R(1+e)}{1-e\sin\theta}\). This means that R(1 + e) = ed. Remember that d is the distance from the focus to the conventional directrix. This allows us the state the distance from the vertex of the conic to the conventional directix in terms of e and R. Specifically, that distance is R/e. This is no surprise since we can also obtain the same result using a, b, and c.

Negative Eccentricity

Figure 3: Negative values of e and conics (to be loaded)

When e < –1, the conic is a hyperbola in blue. When –1 < e < 0, the conic is an ellipse in red. The generating circle is shown in green. The line y = 2R is in cyan.

Other Forms

Originally, when I worked this out, I used a circle of radius R centered at (0, R) and the line y = 2R to derive the equations. Below is what I discovered originally before I reworked out the equations presented above.

The Parabola

A parabola is a locus of all points such that a point on the parabola is equidistant from a given line and a fixed point, the focus. This is the classic definition of a parabola. Its eccentricity, e, is 1.

I have come up with an alternative definition of a parabola, which is similar to the above but involves a circle and a line. A parabola is a locus of all points such that a point on the parabola is equidistant from a fixed line y = 2R and the closest point on the circle with the equation \(y=R\pm \sqrt{R^2-x^2}\). Using this definition, the parabola generated is shown in the Geogebra activity below. You can drag point A and see that the ratios are the same.

The equation of the parabola is given by the equation \(y=2R-\frac{x^2}{4R}\). We can essentially use any tangent line to the circle to generate slanted parabolas.

The parabola in red was generated from the circle in blue and the line y = 2R in light cyan. A vertical segment drawn from the line y = 2R to the parabola is equal in length to a segment drawn from the parabola to a point on the circle such the tangent line at the point is orthogonal to the segment.

The focus of the parabola is the center of the circle or (0, R) — which is what we obtain using the formula 1/(4a) to find the distant of the focus from its vertex for the parabola y = ax².

The directrix of the parabola is the line y = 3R.

The Ellipse and the Hyperbola

If the locus of the points is not equidistant but rather occurs in a ratio of e, then the general equation of the locus of P(x, y) is given by:

(i) \(x^2+(1-e^2)y^2+2R(2e-1)(e+1)y-\) \(4eR^2-4e^2R^2=0\)

or

(ii) \(\frac{x^2}{1+e}+(1-e)y^2+\) \(2R(2e-1)y-\) \(4eR^2=0\)

Equation (ii) is appropriate for a parabola because it reduced to the equation stated above.

The equation in (ii) can be put in graphing form:

(iii) \(y=\frac{-R(2e-1)}{1-e} \pm \) \( \frac{\sqrt{R^2(2e-1)^2+(1-e)(4eR^2-x^2)}}{1-e}\), for which e ≠ 1

The discriminant, D, of a general conic is given by B² – 4AC. The discriminant of (i) is D = 0 – 4(1 – e²) = 4e² – 4.

After completing the square, equation (ii) becomes equisite.

Actually, the above is not the standard form because the right side is not 1. But it looks better that way. In actual standard form, it would be: \(\frac{x^2}{\frac{1+e}{1-e}R^2}+\frac{\left(y+\frac{(2e-1)}{1-e}R\right )^2}{\frac{1}{(1-e)^2}R^2}=1\).

Also note that the above equation can give us a hyperbola is e is > 1, in which case the denominator of the x² becomes negative and the hyperbola opens up and down. Both the ellipse and hyperbola determined by the above equation have the y-axis as the major axis.

From equation (i), when e = 1, D = 0 and the conic is a parabola.

When |e| > 1, the conic is a hyperbola and when 0 < |e| < 1, the conic is an ellipse.

The number e takes upon the classic definition of eccentricity.

Our conic in (i) allows negative values of e. These ranges allow us to cover all shapes of ellipses and hyperbolas that would not be possible if e was restricted to positive values.

When e > 1, the conic is a hyperbola in blue. When 0 < e < 1, the conic is an ellipse in red. The generating circle is shown in green. The line y = 2R is in cyan. (Image to be posted)

A Special Ellipse

When e = ½, the equation reduces a special ellipse: \(\frac{x^2}{3R^2}+\frac{y^2}{4R^2}=1\). This is a great example where we can show that ratio we used here is in fact e as defined in the classic way. Let’s calculate c from a and b in this ellipse. For an ellipse, the following relationship is known: \(c^2 = a^2 - b^2\). For our ellipse, a² = 4R² and b² = 3R². Substituting:

(i) \(c^2=4R^2-3R^2 = R^2\)

(ii) \(c=R\)

The classic formula for e is c/a. Therefore, \(e=\frac{R}{\sqrt{4R^2}} = \frac{R}{2R} = \frac{1}{2}\). This agrees with our definition of the eccentricity!!

Why is the ellipse with e = ½ special? Well, first, its center is at the origin (0, 0). Second, its foci are located at (0, R) and (0, –R). One of the focus points is the center of the circle that we used to generate the conic. The center of this circle always seem to be one of the foci of our conics. Moreover, the foci are the midpoints of the half of the major axis.

The Geogebra activity above shows an ellipse generated from a circle and a line. You can drag the point P around and see that the ratio is contant. You may also change the radius of the circle and the eccentrity of the of ellipse by using the sliders.

The Special Hyperbola

Just as we found a special ellipse above, we have a special hyperbola when e = 3/2. The equation is: \(-\frac{x^2}{5R^2} + \frac{(y+3R)^2}{4R^2}\). In the hyperbola, \(a = \sqrt{5}R\), \(b = 2R\), and \(c = 3R\).

The center of the hyperbola is (0, –3R). One focus is of course the center of the circle, which is (0, 0), The other focus is at (0, –6R). The vertices are (0, –R) and (0, –5R).

The asymptotes are \(y=\pm\frac{2}{\sqrt{5}}x-3R\). The hyperbola is pictures below.

The Circle

The circle has an eccentricity of 0. However, when we define the conic using a polar function using a line (the directrix) and a point (the focus), the equation we obtain does not allow eccentricity to be 0. When the polar equation is converted into a rectangular form, letting e = 0 gives a degenerate conic - a point.

Our equation above actually gives us an equation of the circle - the same circle that generated other conics. We can let e = 0 in the equation \(x^2+(1-e^2)y^2+\) \(2R(2e-1)(e+1)y-\) \(4eR^2-\) \(4e^2R^2=0\) and see that it reduces to the same circle, which essentially means the ratio of the distance from the line to the circle to the closest point on the circle is 0.

(i) \(x^2+(1)y^2+2R(1)(1)y=0\)

Completing the square:

(ii) \(x^2+(y-R)^2=R^2\)

And there you have it. An eccentricity of 0 gives us a circle.

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