SME
Grade 12 Math · Problem type
Polar conics with sin θ
Use this for “Name the conic r = 4 ÷ (1 + sin θ), its eccentricity and its directrix.”
Diagram
The
equation
r
=
4
1 + sin θ
gives
e
=
1
and
ed
=
4,
so
d
=
4.
With e = 1 the curve is a parabola: its focus is the pole and its directrix is y = 4.
The point at θ = 90° has r = 2.
The
point’s
distance
to
the
focus
is
PF
=
2
and
to
the
directrix
PD
=
2,
and
PF
PD
=
1,
which
is
e.
This is an example. Type your own values to start a new problem.
r
=
k
−
(
)
sin
°
e
Eccentricity
calculated
d
Distance from the focus to the directrix
calculated
Formulas
e
=
|
n
|
÷
m
1 = |−1| ÷ 1
d
=
k
÷
|
n
|
4 = 4 ÷ |−1|
r
=
k
m
−
n
× sin(
θ
)
2
=
4
1 − (−1) × sin(90°)
Assumptions
•
Divide the top and bottom by m to reach r = ed ÷ (1 − e sin θ): the eccentricity e and the directrix distance d.
•
e < 1 is an ellipse, e = 1 a parabola, e > 1 a hyperbola; the focus is at the pole.
•
n is the number after the minus: + sin θ is n = −1. The directrix is horizontal: y = −d with − sin θ, y = d with + sin θ.
Step-by-step
WE KNOW
k = 4, m = 1, n = −1, θ = 90°
FIND
eccentricity (e), distance from the focus to the directrix (d), distance from the pole (r)
Step 1 · Find eccentricity (e)
Use e = |n| ÷ m
Dividing
the
bottom
by
m
leaves
1
−
e
sin
θ
,
so
e
is
the
number
before
sin
θ
over
m
.
e
=
|−1|
÷
1
e
=
1
→
a
parabola
Step 2 · Find distance from the focus to the directrix (d)
Use d = k ÷ |n|
The
top
over
m
is
ed,
and
e
is
|
n
|
÷
m
,
so
d
=
k
÷
|
n
|.
d
=
4
÷
|−1|
d
=
4
→
the
directrix
is
y
=
4
Step 3 · Find distance from the pole (r)
Use r = k ÷ (m − n × sin(θ))
Put the angle into the equation: the point on the conic in that direction.
r
=
4
1 − (−1) × sin(90°)
r
=
4
1 − (−1)
r
=
4
2
r
=
2
Check
Put the numbers back in. Both sides should match.
1 = |−1| ÷ 1 ✓
4 = 4 ÷ |−1| ✓
2
=
4
1 − (−1) × sin(90°)
✓
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