Conic Section: Parabola in Rectangular, Polar, and Parametric Forms

Rectangular: In the image at the right, you have a parabola with its focus at the origin. Recall the definition of a parabola is the set of all points in the plane of a a fixed point (the focus) and a fixed line (the directrix) that are equidistance from the focus and directrix. From this definition, you can derive the equation for the pictured parabola as y² = 4p(x + p). Note that in this image p stands for the distance from the vertex to the focus which is equal to the distance from the vertex to the directrix. Hence p is positive. Also, in general p is considered positive when the parabola opens in a positive direction.

Polar: It is easy to derive the equation in polar coordinates, since we have the focus at the origin. The definition of the parabola says that the distance from an arbitrary point on the curve to the focus must be equal to the distance from the directrix, and we call the distance from the origin to the point on the curve r, then the distance from the point to the directrix must also be r as illustrated by the horizontal line segment near the top of the image. By comparing to the projection of that line segment onto the x-axis, you can see that r must equal 2p + rcos θ. Hence, we have the polar equation: r = 2p + rcos θ.

Parametric: To go from the polar equation to a set of parametric equations, just convert using x = rcos θ and y = rsin θ. Thus we need to solve the polar equation for r. We get r = 2p/(1 - cos θ), so x = 2p cos θ/(1 - cos θ) and y = 2p sin θ/(1 - cos θ). Pick any value for p and let θ range for 0 to 2π, then graph this using David Little's Parametric Equation Investiagor and you will see that you get a parabola. You will also see that the graph of x versus θ appears to be the absolute value or square of cosecant and the graph of y versus θ appears to be a transformed cotangent function. With a bit of algebraic manipulation and letting p = 1, you can show that x = csc²(θ/2) - 2 and y = 2 cot (θ/2). Of course, there are many other ways to get a parabola and we don't have to keep the origin at the focus so it is a good exercise to play with the parametric equations, then look to see which pairs of equations appear to give you a parabola, then verify that you indeed have a parabola algebraically by converting back into a rectangular equation. You can also use trig identities to further manipulate the parametric equations given here.

Exercises: For numbers 1-11, show the set of parametric equations result in a parabola by converting the equation to a rectangular equation and identify the vertex, the value of p, and the direction that the parbola opens. Also state the minumum domain needed for t to create the complete parabola.

1.) x = csc2(t/2) - 2      y = 2 cot (t/2)	2.) x = csc2(t) - 2      y = 2 cot (t)	3.) x = csc2(t)      y = cot (t)	4.) x = 4 sec2(t)      y = 6 tan (t)
5.) x = cot (t)      y = csc2 (t)	6.) x = 2 tan t      y = sec2t	7.) x = tan t      y = sec2 t - 1	8.) x = cot t      y = csc2 t - 1
9.) x = tan t      y = tan 2t	10.) x = tan t + 7        y = tan2 t - 4	11.) x = tan 2(t) + 3        y = tan (t) + 1	12.) Write (x - h)2 = 4p(y - k) as a set of parametric equations where -π/2 ≤ t ≤ π/2.