r = ep
or
1 + e cos q
r = ep
1 + e sin q
e > 0 is the eccentricity and |p| is the
distance between the pole and the directrix.
If the denominator is 1 + e sin q,
it has a horizontal directrix above the pole.
If the denominator is 1 - e sin q,
it has a horizontal directrix below the pole.
If the denominator is 1 + e cos q,
it has a vertical directrix to the right of the pole.
If the denominator is 1 - e cos q,
it has a vertical directrix to the left of the pole.
Examples:
(a) Sketch
r =
3
4 + 2 sin q
(i) Which conic section is it? Divide so the first number in the denominator is 1. Check e.
Divide everything in the fraction by 4 so the equation will
be in the correct form to pick out
the parts.
r
= 3/4
1 + (1/2) sin q
The
graph has a horizontal directrix above the pole.
e
= 1/2 so the graph is an ellipse.
ep
= 3/4 means 2p = 3/4 so
p = 3/8
Therefore,
the graph is an ellipse with directrix 3/8 units above the
pole (a focal point).
(ii) Use e = c/a
and a2 = b2 + c2
or a2 + b2 = c2
At q = p/2,
r = 3/4
= 3/4 = 1/2
1 + 1/2 3/2
So the graph hits (p/2, 1/2).
At
q = 3p/2, r
= 3/4
= 3/4 = 3/2
1 - 1/2 1/2
So the graph hits (3p/2, 3/2).
At q = p, r
= 3/4
= 3/4 = 3/4
1 - 0
1
So the graph hits (p, 3/4).
At q = 0, r =
3/4 = 3/4 =
3/4
1 - 0 1/2
So the graph hits (0, 3/4).
(b) Find the equation:
(i) e = 1;
directrix x = -1
The graph has a vertical directrix, so the equation will use cos
q.
|p| = 1 (the distance between the pole and directrix)
So
r = 1
1 - cos q
(ii) e = 3/4; directrix y = -2
The graph has a horizontal directrix, so the equation will use sin
q.
|p| = 2 (the distance between the pole and directrix)
r =
3/2 =
6
1 - (3/4) sin q
4 - 3 sin q
Problems (there are no problems here yet)
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