The sum of the two distances never changes.
Pin a loop of string to two nails and pull it taut with a pencil: the curve you trace is an ellipse, because the string length never changes.
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- The Ellipse
The Ellipse
Two foci, and the two distances always add to the same total.
Find the foci
x²25 + y²9 = 1
The larger denominator is under x², so the major axis lies along the x-axis with a² = 25 and b² = 9. Then c² = a² − b² = 16, so c = 4 and the foci are (±4, 0). A puts them on the wrong axis. B gives the vertices, not the foci. C adds the denominators — that is the hyperbola's rule, not the ellipse's.
The larger denominator sits under the major axis.
x²100 + y²36 = 1
major axis horizontal
a = 10, b = 6
x²36 + y²100 = 1
major axis vertical
a = 10, b = 6
a is always the larger
a > b
by definition, not by position
a belongs to the bigger number, wherever it appears
Assuming a² is always under x² is the most reliable way to get every later answer wrong. Look at which denominator is larger first, and put the foci on that axis.
For an ellipse, c² = a² − b².
a² = 100, b² = 36
larger is a²
c² = 100 − 36 = 64
SUBTRACT for an ellipse
c = 8
foci (±8, 0)
Because c is smaller than a, the foci always sit inside the ellipse, on the major axis. If your c comes out bigger than a, you have added instead of subtracted.
A shifted centre changes nothing else.
centre (−3, 2)
signs flip as always
a² = 16, b² = 4
horizontal major axis
c² = 12, c = 2√3
measured from the centre
foci (−3 ± 2√3, 2)
along the major axis
Vertices are a units from the centre along the major axis; co-vertices are b units along the minor. Everything is measured from the centre, never from the origin.
A point on the curve: substitute the one coordinate you know.
Arches, tunnels and orbits ask for a height at a given distance from the centre. The equation answers it directly.
How high is x²100 + y²25 = 1 above the x-axis at x = 6?
36100 + y²25 = 1
put x = 6 in
y²25 = 1 − 36100 = 64100
move the known part across
y² = 25 × 64100 = 16
multiply by b²
y = 4
the height is the positive root
Measure x from the CENTRE, not from one end of the arch. Here the ends of the curve are at x = ±10, so x = 6 is 6 units out from the middle.
An arch given only by its size builds the same equation: a semi-elliptical arch 20 m wide and 5 m high has a = half the width = 10 and b = the height = 5, which is exactly x²100 + y²25 = 1, so 6 m from the centre it stands 4 m high.
Variations: shifted and vertical, not equal to 1, and backwards.
The same three moves: find a², find c from c² = a² − b², step from the centre.
Shifted and vertical: (x − 1)²20 + (y + 2)²36 = 1
centre (1, −2); 36 is larger, under y²
vertical major axis
c² = 36 − 20 = 16, c = 4
subtract
foci (1, 2) and (1, −6)
4 up and 4 down from the centre
On a vertical ellipse the foci share the centre's x-coordinate; only y moves.
Not equal to 1 yet: 9x² + 16y² = 144
x²16 + y²9 = 1
divide every term by 144
c² = 16 − 9 = 7
a² = 16 under x²
foci (±√7, 0)
an irrational c is fine
Divide by the constant first; reading 9 and 16 as the denominators swaps the axes.
Backwards: vertices (0, ±13), foci (0, ±12)
a = 13, c = 12, vertical
both pairs on the y-axis
b² = a² − c² = 169 − 144 = 25
rearrange c² = a² − b²
x²25 + y²169 = 1
a² goes under y²
Working backwards, b² is what we solve for; it still comes from subtracting.
Eccentricity measures how squashed it is.
e = ca
always between 0 and 1
for an ellipse
e near 0
almost a circle
foci nearly together
e near 1
long and thin
foci near the ends
A circle is an ellipse with e = 0, where the two foci have merged into the centre and a = b. The Earth's orbit has e ≈ 0.017, which is why it looks circular in every diagram even though it is not.
On the exam
Tests give an equation and ask for the foci, vertices or the major axis, or ask for a height on an elliptical arch. The fastest route: put a² under the larger denominator, subtract for c², and step c and a from the centre along that axis. A 10-second check: c must come out smaller than a, so the foci land inside the vertices.
Now you try
Practice problem
Find the foci
x²9 + y²25 = 1
The larger denominator is now under y², so the major axis is VERTICAL with a² = 25 and b² = 9. Then c² = 16 and the foci are (0, ±4). B keeps them on the x-axis out of habit. C gives the vertices.
Hint: the 25 has moved. Check which denominator is larger before doing anything else, because that decides the axis the foci sit on.