SME
Grade 12 Math · Problem type
An ellipse traced by an angle
Use this for “Eliminate the parameter from x = 1 + 3 cos t, y = −2 + 2 sin t.”
Diagram
At t = 60° the point is (2.5, −0.27).
The arrows show the way t runs.
This is an example. Type your own values to start a new problem.
x
=
h
+
a
cos
t
°
y
=
k
+
b
sin
t
°
Formulas
x
=
h
+
a
×
cos(
t
°)
2.5 = 1 + 3 × cos(60°)
y
=
k
+
b
×
sin(
t
°)
−0.2679 = −2 + 2 × sin(60°)
Assumptions
•
t is an angle from 0° to 360°; the point goes around once, counterclockwise.
•
cos t = (x − h) ÷ a and sin t = (y − k) ÷ b, and cos²t + sin²t = 1.
•
So (x − h)²/a² + (y − k)²/b² = 1: an ellipse with center (h, k), a circle when a = b.
Step-by-step
WE KNOW
t = 60°, h = 1, k = −2, a = 3, b = 2
FIND
x at t (x), y at t (y)
Step 1 · Find x at t (x)
Use x = h + a × cos(t°)
Start
at
the
center
and
go
a
cos
t
across.
x
=
1
+
3
×
cos(60°)
x
=
1
+
3
×
0.5
x
=
1
+
1.5
x
=
2.5
Step 2 · Find y at t (y)
Use y = k + b × sin(t°)
Start
at
the
center
and
go
b
sin
t
up.
y
=
−2
+
2
×
sin(60°)
y
=
−2
+
2
×
(
√
3
2
)
=
−2
+
√
3
y
=
−0.2679
→
(
x
− 1)
2
/9
+
(
y
+ 2)
2
/4
=
1
Check
Put the numbers back in. Both sides should match.
2.5 = 1 + 3 × cos(60°) ✓
−0.2679 ≈ −2 + 2 × sin(60°) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.