Question

explain Pole and Polar of a Parabola, graphically
(the locus of point of intersection of tangents drawn at extremities of chord)

Answer 1

Let F be a fixed point and l a fixed line in the plane. For any point P consider the two distances:



Source - http://colalg.math.csusb.edu/~devel/IT/main/m11_conic/src/s08_polar-conics.html

d(P, F) - the distance between P and F
d(P, l) - the distance between P and l:

We are interested in the ratio ; it is used to define the conic sections as follows:

A collection of points P in the plane such that e = is a fixed positive number is called a conic section. The number e is called the eccentricity of the conic. The line l is called the directrix of the conic, and the point F is called the focus of the conic.

If 0 < e < 1, then the conic is an ellipse
If e = 1, then the conic is a parabola
If e > 1, then the conic is an hyperbola

We have already seen the parabola defined in terms of a directrix and a focus. This definition shows that ellipses and hyperbolas can also be defined in terms of a directrix and focus.

If we position the point F at the pole and choose a directrix to be either a line parallel or a line perpendicular to the polar axis, then the polar equation of a conic turns out to have a fairly simple form.

Consider the point F located at the pole and the directrix, l, a vertical line with Cartesian equation x = d, d > 0. Let the point P have Cartesian coordinates (x,y) and polar coordinates (r,).
The left side of the equation d(P,F) = ed(P,l) is simply r, while the right side is e(d − rcos()). Hence,
r = ed − ercos())
Solve for r and obtain
r = ed/(1 + ecos())


Had we chosen the directrix to be the vertical line with Cartesian equation x = −d (so the directrix would be to the left of the pole), we would have found the equation of the conic to be
r = ed/(1 − ecos())

You might like to verify that this is indeed the equation. We obtain a similar equation if we take the directrix to be parallel to the polar axis. For example, if the directrix is the horizontal line with Cartesian equation y = d, d > 0, we get the equation
r = ed/(1 + esin())

We summarize these results.

Let d be a positive number. Four conics of eccentricity e with focus at the pole are:
directrix x = d r = ed/(1 + ecos())
directrix x = −d r = ed/(1 − ecos())
directrix y = d r = ed/(1 + esin())
directrix y = −d r = ed/(1 − esin())

In the following demonstration, you can see the graphs of
r = ed/(1 ecos()), and r = ed/(1 esin())
for differing values of e and d. Click on the button labeled "sin" to change to the cosine function. Click on the button labeled "+" to change the sign preceding the trigonometric function. You will be shown the conic defined by the values you have selected, and the directrix. Remember that the point F is always chosen to be at the pole.

Answer 2

thank you! but the above description sounds like polar equations