SME
Grade 12 Math · Precalculus
Conic sections: parabolas, ellipses and hyperbolas
Use this for “Find the center, vertices, foci and eccentricity of the ellipse (x − 1)²/25 + (y + 2)²/9 = 1.”
Diagram
(x − 1)²/25 + (y + 2)²/9 = 1
Center (1, −2)
c
=
√
25 − 9
=
4:
the
foci
are
4
from
the
center,
across.
From any point, the distances to the two foci add to the long axis.
This is an example. Type your own values to start a new problem.
(x
−
)²
a
2
+
(y
−
(
)
)²
b
2
=
1
c
Center to each focus
worked out (not typed)
e
Eccentricity
worked out (not typed)
Formulas
c
=
√
|
a
2
−
b
2
|
4
=
√
|
5
2
−
3
2
|
e
=
c
max(
a
,
b
)
0.8
=
4
max(5, 3)
Assumptions
•
a goes with x and b with y: a > b is a wide ellipse, a < b a tall one.
•
The foci lie on the long axis, c from the center, with c² = a² − b² (the larger square first).
•
From any point on the ellipse, the distances to the two foci add to the long axis, 2a or 2b.
Step-by-step
WE KNOW
h = 1, k = −2, a = 5, b = 3
FIND
center to each focus (c), eccentricity (e)
Step 1 · Find center to each focus (c)
Use c = √(|a² − b²|)
The
long
half-axis
squared
is
c
2
plus
the
short
one
squared:
take
the
smaller
square
from
the
larger.
c
=
√
a
2
−
b
2
c
=
√
5
2
−
3
2
c
=
√
25 − 9
c
=
√
16
c
=
4
→
foci
(−3,
−2)
and
(5,
−2);
vertices
(−4,
−2)
and
(6,
−2)
Step 2 · Find eccentricity (e)
Use e = c ÷ max(a, b)
Eccentricity
is
c
over
the
long
half-axis:
0
is
a
circle,
near
1
very
flat.
e
=
c
a
e
=
4
5
e
=
0.8
Check
Put the numbers back in. Both sides should match.
4
=
√
|
5
2
−
3
2
|
✓
0.8
=
4
5
✓
Topic check
See what you know: 5 questions on this topic
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Conics as slices of a cone
Use this for “Why are circles, ellipses, parabolas and hyperbolas called conic sections?”