**CONIC SECTIONS IN POLAR COORDINATES**

**PART I - CONIC SECTIONS REVIEW**

**(1) PARABOLA:**

In Assignments 2 and 3, we learned that if P is any point on the parabola, F is the focus, and D is the directrix, then the distance PF is equal to the distance PD. Therefore the ratio PF/PD=1. Let's give a name to this ratio: eccentricity. PF/PD=k. Parabola is a special conic section with eccentricity k = 1. See figure below:

**(2) ELLIPSE:**

Consider an ellipse with foci F(c,0) and F'(-c,0), and vertices on the x-axis A(a,0) and A'(-a,0). (We must remind our students that foci is the plural of focus and vertices is the plural of vertex). Let P(x,y) be any point on the ellipse. Then by definition of the ellipse, the sum of distances from P to foci is 2a. Namely, PF+PF'=2a. Have you heard of diretcrix of an ellipse? For consistency, let us define a directrix via the equation x = a^2 / c. See figure below for the derivation of the ratio PF/PD for the ellipse:

**(3) HYPERBOLA:**

Consider an hyperbola with foci F(c,0) and F'(-c,0), and vertices A(a,0) and A'(-a,0). Let P(x,y) be any point on the hyperbola. Then by definition of the hyperbola, the difference of distances from P to foci is 2a. Namely, PF-PF'=2a. Again for the sake of consistency, let us define a directrix via the equation x = a^2 / c. See figure below for the derivation of the ratio PF/PD for the hyperbola (It's very similar to what we did for ellipse. The only difference is that the eccentricity k > 1 now.)

**PART II - POLAR EQUATIONS OF CONICS**

If a conic section with eccentricity k is positioned in a polar coordinate system so that its focus is at the pole and the corresponding directrix is p units from the pole and is either parallel or perpendicular to the polar axis, then the equation of the conic has one of four possible forms, depending on its orientation:

**CASE ONE: DIRECTRIX RIGHT OF POLE**

If we fix p and animate k between 0 and say, 2, we can observe all conic sections for which the directrix is p units to the right of the pole at once.

Click the figure below to download the animation of all conic sections (with trace enabled) together:

**CASE TWO: DIRECTRIX LEFT OF POLE**

**CASE THREE: DIRECTRIX ABOVE POLE**

Does it make sense? It looks like we rotated the conic sections! In fact, there is nothing ambiguous here. Just remember the relation between sine and cosine functions. Remember from the first assignment that sine and cosine functions are 90 degrees out of phase. Mathematically, cos (X) = sin ( x + Pi / 2). And this refers to a CLOCKWISE ROTATION when we switch from CASE TWO to CASE THREE above. It therefore makes sense and there is nothing ambiguous here.

**CASE FOUR: DIRECTRIX BELOW POLE**

**PART III - APPLICATIONS IN ASTRONOMY - KEPLER'S
LAWS**

FIRST LAW: All planets move in elliptical orbits with the Sun at one of the focal points (FOCI) of the ellipse. (LAW OF ORBITS)

SECOND LAW: The radius vector drawn from the Sun to any planet sweeps out equal areas in equal time intervals. (LAW OF AREAS)

THIRD LAW: The square of the orbital period
of any planet is proportional to the cube
of the semimajor axis of the elliptical orbit.
(LAW OF PERIODS)

BODY | T = PERIOD (10^6 s) | A = SEMIMAJOR AXIS OF THE ELLIPSE (10^10 m) | T^2 / A^3 RATIO (10^-19) | ECCENTRICITY |

MERCURY | 7.6 | 5.79 | 2.97 | 0.206 |

VENUS | 19.4 | 10.8 | 2.99 | 0.007 |

EARTH | 31.56 | 14.96 | 2.97 | 0.017 |

MARS | 59.4 | 22.8 | 2.98 | 0.093 |

JUPITER | 374 | 77.8 | 2.97 | 0.049 |

SATURN | 935 | 143 | 2.99 | 0.051 |

URANUS | 2640 | 287 | 2.95 | 0.046 |

NEPTUNE | 5220 | 450 | 2.99 | 0.005 |

PLUTO | 7820 | 591 | 2.96 | 0.250 |

QUESTION: Which planet has the most nearly circular orbit?

ANSWER: Neptune has the smallest eccentricity, therefore, it is the planet with the most nearly circular orbit.

The terms *perihelion *and *aphelion *are used to denote the position of a planet
around the Sun. The **perihelion** is the point nearest the Sun; the **aphelion** is the point farthest from the Sun. The
length of the semimajor axis of a planet's
elliptical orbit is called the **mean distance **of the planet from the Sun.

ANSWER: The mean distance, namely the length of the semimajor axis of the ellipse a=884.74 million miles. Once we figure out how to find the length of the semiminor axis of the ellipse b, we are done. Let's locate the center of this ellipse at the origin of the Cartesian coordinates. We can find how far the Sun is from the center of the ellipse = 934.34 - 884.74 = 49.6 million miles. Then the Sun will be at the focus (-c, 0) = (-49.6 million miles, 0).

We now know a and c, therefore we can get b using the equation above: b=883.35 million miles.

Therefore the equation of the ellipse in standard form is:

x^2 / a^2 + y^2 / b^2 = 1, where, a = 884.74 x 10^6, and, b=883.35 x 10^6

**PART IV - APPLICATIONS IN ARCHITECTURE -
COLOSSEUM**

The base of the Colosseum in Rome has an elliptical shape. The length of the major axis is 188 meters and the mionr axis is 156 meters (approximately).

ANSWER: