UNIT 3: ELLIPSE
At the end of the lesson, you should be able to:
(1) define an ellipse;
(2) recognize the general equation and important characteristics of an ellipse;
(3) determine the standard form of equation of an ellipse; and
(4) solve situational problems involving ellipse.
ENGAGE
Whispering Galleries -- Mormon tabernacle
The Mormon Tabernacle in
Salt Lake City has an
elliptical ceiling. You can
hear a pin drop from 175
feet away. The Tabernacle
is 250 feet long, 150 feet
wide, and 80 feet high. The
organ has 11,623 pipes
Statuary Hall in the U.S. Capital
The Hall was constructed in the shape of an ellipse.
It is said that if you stand at one focal point of the
ellipse, you can hear someone whispering across the
room at the other focal point because of the
acoustical properties of the elliptical shape. The
gallery used to be a meeting place of the House of
Representatives. According to legend, it was John
Quincy Adams that discovered the room’s sound
properties. He placed his desk at a focus so he could
easily hear conversations across the room.
➢ The YouTube video illustrates this
phenomenon.
https://youtu.be/FX6rUU_74kk
➢ You may view this also in your memory stick. Whispering gallery – National Statuary
Hall
EXPLORE
Do this: In this activity, you will need a clean sheet of paper, ruler, string,
pins, and a pencil.
❖ here are some links for this activity:
https://www.mathopenref.com/constellipse1.html
https://www.youtube.com/watch?v=0maahsJQOJE
After doing this
Start with the height and width of the
desired ellipse. The two lines are
the major and minor axes of the ellipse.
The major axis is the longer one.
1. With the compasses' point on the
center, set the compasses' width to half
the width (major axis) of the desired
ellipse.
You may also use a ruler to measure this
length from the center to one vertex.
(This is called the ellipse semi major
axis).
2. Move the compasses' point to one end
of the minor axis of the desired ellipse and
draw two arcs across the major axis.
Using your ruler, measure the same
length from one end of the minor axis to
the major axis.
Your work should look like this
After doing this
3. Where these arcs cross the major axis
are the foci of the ellipse. Label them F1,
F2.
4. Put a pin in each end of the major axis
(they will be moved later), and tie a string
to them so that the string between them is
taut. The best way to do this is to push the
pin through the string itself if possible,
rather than tying a knot.
5. Leaving the string attached, move the
pins to the focus points F1, F2. Put a
pencil point against the string and pull
the string taut with the pencil.
6. Keeping the string taut, move the
pencil in a large arc. The pencil will draw
out the desired ellipse. To avoid the string
catching on the pins, you may find it better
to draw the upper and lower halves of the
ellipse separately.
Your work should look like this
After doing this
Your work should look like this
7. Done. The ellipse will pass through the
four initial points defining the ends of the
major and minor axes.
8. Label the ends of the major axis as V1
and V2. Locate at least two more points
on any part of the graph.
P1
V2
V1
P2
Measure the distance between vertices |𝑉1𝑉2| = 2a = ______
The distance between the center and vertex is a therefore, a = ______
Measure |𝐹1𝑃1| + |𝐹2𝑃1| = ________
Measure |𝐹1𝑃2| + |𝐹2𝑃2| = ________
What is the relationship between the sum of the distances between points P to the foci and
the distance between vertices?
From this activity, how do you define an ellipse?
Definition of an ellipse as a locus of points
An ellipse is a set of points in a plane such that the sum of the
distances from two fixed points called the foci (plural of focus) is constant.
And, for any point on the ellipse,
|𝑃𝐹1 | + |𝑃𝐹2 | = 2𝑎
where a is the distance between the center and one vertex.
Terms related to an ellipse
1. Foci – the two fixed points (F1 & F2)
2. Principal axis – the line passing through the foci
3. Vertices – the points of intersection of the ellipse and its principal axis
4. Major axis – the line segment whose endpoints are the vertices of the ellipse
5. Minor axis – the line segment through the center perpendicular to the major axis with end
points on the ellipse
Note: the major axis is always longer than the minor axis
6. Center – the midpoint of the major axis
7. Co – vertices – endpoints of the minor axis (some books also call this the pseudo vertices)
8. Focal distance – the distance from the center to the focus
Important distances:
𝒄𝟐 = 𝒂𝟐 − 𝒃𝟐
a = the distance between the center and
the vertex; half of the length of the major
axis
b = the distance between the center and
the co - vertex; half of the length
of the minor axis
c = the distance from the center to the
focus
note: a > b, the major axis is longer
than the minor axis
Equation of an ellipse with center at the origin
Major axis
Ellipse
Center Foci
Vertices
Horizontal
( 0, 0 )
( c, 0 )
( a, 0 )
( -c, 0 )
( -a, 0 )
Vertical
( 0, 0 )
( 0, c )
( 0, -c )
( 0, a )
( 0, -a )
Minor axis
Co-vertices
( 0, b )
( 0, -b )
( b, 0 )
( -b, 0 )
Equation
𝒙 𝟐 𝒚𝟐
+
=𝟏
𝒂𝟐 𝒃𝟐
𝒚𝟐 𝒙 𝟐
+
=𝟏
𝒂𝟐 𝒃𝟐
➢ The foci of the ellipse are on the major axis.
➢ In the standard equation of ellipse, if the x- part has the bigger denominator, the ellipse is
horizontal.
➢ In the standard equation of ellipse, if the y- part has the bigger denominator, the ellipse is
vertical.
Examples:
1. Give the coordinates of the foci, vertices, and co-vertices, and focal length of the ellipse
a.
𝑥2
+
25
𝑦2
9
=1
b. 4𝑥 2 + 𝑦 2 = 16
Solution:
a. 𝑎2 = 25; 𝑎 = 5
𝑏2 = 9 ; 𝑏 = 3
𝑐 2 = 𝑎2 − 𝑏 2 = 25 − 9 = 16; 𝑐 = 4
Foci (-4, 0) (4, 0)
Vertices (-5, 0) (5, 0)
Co-vertices (0, 3) (0, -3)
b. 4𝑥 2 + 𝑦 2 = 16
Since the right side of the equation of an ellipse is
always equal to 1, we then multiply the equation by
1
16
1
(4𝑥 2 + 𝑦 2 = 16) ( )
16
𝑥2 𝑦2
+
=1
4 16
𝑎2 = 16; 𝑎 = 4
𝑏2 = 4 ; 𝑏 = 2
𝑐 2 = 𝑎2 − 𝑏 2 = 16 − 4 = 12; 𝑐 = 2√3
Foci (0, 2√3) (0, −2√3 )
Vertices (0, 4) (0, - 4)
Co-vertices (2, 0) (- 2, 0)
2. Find the equation of the ellipse with center at the origin given the following properties:
a. with one vertex at (0, 5) and one focus at (0, -3)
Solution: distance from the origin to the vertex (0, 5)
a=5
distance from the origin to the focus (0, -3)
c=3
𝑐 2 = 𝑎2 − 𝑏 2 ; 𝑏 2 = 𝑎2 − 𝑐 2
𝑏 2 = 52 − 32 = 16; 𝑏 = 4
Since the vertex and the focus are both on the
y – axis, the ellipse is a vertical ellipse with
𝑥2
𝑦2
+ 2=
2
𝑏
𝑎
𝑥2
𝑦2
+ =1
16
25
standard equation
by substitution:
1;
b. the focus and the co-vertex are 2 units from the center with x – axis as the principal axis
Solution: From the given data, this is a horizontal ellipse with coordinates of the foci (∓ 2, 0) and
the coordinates of the co-vertices (0, (∓ 2).
distance from the origin to one
co-vertex (0, 2) b = 2
distance from the origin to the
focus (2, 0)
c=2
𝑐 2 = 𝑎2 − 𝑏 2 ; 𝑎2 = 𝑏 2 + 𝑐 2
𝑎2 = 22 + 22 = 8; 𝑎 = 2√2
Since the principal axis is the x-axis, the
ellipse is a horizontal ellipse with
𝑥2
𝑦2
+ 2=
2
𝑎
𝑏
𝑥2
𝑦2
+ =1
8
4
standard equation
by substitution:
1;
Now, let as shift the center to an arbitrary point (h, k) on the plane.
Equation of an ellipse with center at (h, k)
Major axis
Ellipse
Center Foci
vertices
(h - c, k)
Horizontal
Equation
Co-vertices
(h, k + b)
(h + c, k)
(h - a, k)
(h + a, k)
(h, k + c)
(h, k + a)
(h + b, k)
(h, k - c)
(h, k - a)
(h - b, k)
(h, k)
Vertical
Minor axis
(h, k - b)
(h, k)
( x − h) 2 ( y − k ) 2
+
=1
a2
b2
( y − k ) 2 ( x − h) 2
+
=1
a2
b2
Examples:
1. Give the coordinates of the foci, vertices, and co-vertices of the ellipse and sketch the graph
with equations
a.
(𝑥+1)2
9
+
(𝑦−2)2
25
= 1 ; 𝐶(−1, 2)
Since a is the denominator under the y
variable, therefore this ellipse is vertical
𝑎2 = 25; 𝑎 = 5
𝑉𝑒𝑟𝑡𝑖𝑐𝑒𝑠(_____,
_____) (_____, ______)
𝑏2 = 9 ; 𝑏 = 3
𝐶𝑜 − 𝑣𝑒𝑟𝑡𝑖𝑐𝑒𝑠 (_____, _____) (______, _____)
𝑐 2 = 𝑎2 − 𝑏 2 = 25 − 9 = 16; 𝑐 = 4
𝐹𝑜𝑐𝑖: (_____, ______) (______, _______)
b.
(𝑥 − 1)²
5
+ (𝑦 + 3)² = 4;
Since the right side of the equation is not equal
1
to 1, we multiply this first by
4
(𝑥 − 1)²
5
Thus, [
(𝑥 − 1)²
20
+
+
(𝑦 + 3)²
1
(𝑦 + 3)²
4
1
4
= 4] ( )
= 1; 𝐶 (_____, ______)
𝑎2 = 20; 𝑎 = 2√5 ≈ 4.47
𝑉𝑒𝑟𝑡𝑖𝑐𝑒𝑠: (_____, _____) (______, _____)
𝑏2 = 4 ; 𝑏 = 2
𝐶𝑜 − 𝑣𝑒𝑟𝑡𝑖𝑐𝑒𝑠 (_____, _____) (______, _____)
2
2
𝑐 = 𝑎 − 𝑏 2 = 20 − 4 = 16; 𝑐 = 4
𝐹𝑜𝑐𝑖: (_____, _____) (______, _____)
2. Find the equation of the ellipse given the following properties:
a. Center at (4, 1); vertex at (4, 4); a focus at (4, 3)
Solution: We know that the center, vertex and focus are all on the principal axis, x = 4.
Thus, this ellipse is vertical.
Solve for the distance from the center to the
vertex, a, and the distance from the center to the
focus, c, by distance formula
𝑎 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
𝑎 = √(4 − 4)2 + (4 − 1)2 = 3
𝑐 = √(4 − 4)2 + (3 − 1)2 = 2
𝑏 2 = 𝑎2 − 𝑐 2 = 32 − 22 = 5;
𝑏 = √5 ≈ 2.24
( 𝑦 − 𝑘 ) 2 (𝑥 − ℎ ) 2
+
=1
𝑎2
𝑏2
(𝑦 − 1)2 (𝑥 − 4)2
+
=1
(_____)
(_____)
b. Endpoints of the minor axis (1, 3) (1, -1); a focus at (-1, 1)
Solution: To find the center, we get the midpoint of the minor axis
𝑥 +𝑥
𝑦 +𝑦
Midpoint formula ( 2 1 , 2 1 )
1+1
(
2
2
3 +(−1)
);
2
2
𝐶
,
𝐶(_____, ______)
Get the distance from center to focus, c, and
center to co-vertex, b, by distance formula
𝑐 = √(1 − (−1))2 + (1 − 1)2 = 2
𝑏 = √(1 − 1)2 + (1 − 3)2 = 2
𝑎2 = 𝑏 2 + 𝑐 2 = 22 + 22 = 8;
𝑎 = √8 ≈ 2.83
(𝑥 − ℎ )2 (𝑦 − 𝑘 )2
+
=1
𝑎2
𝑏2
(𝑥 − 1)2 (𝑦 − 1)2
+
=1
(_____)
(_____)
Situational Problems Involving ellipse
1. The orbit of the Earth around the Sun is in the shape of an ellipse, where the Sun is at one of
the foci, and whose length of the major axis is 185.8 million miles. If the distance of the Sun from
the center of the orbit is 1.58 million miles, then find the least distance and the greatest distance
of Earth from the Sun.
Solution:
2𝑎 = 185.8 ; a = 92.9
𝑐 = 1.58
Answer:
Least distance: S = a – c = 92.9 – 1.58
= 91.32 million miles
Greatest distance: L = a + c = 92.9 +
1.58 = 94.48 million miles
2. The arch of a bridge is in the shape of a semi ellipse, with its major axis at the water level.
Suppose the arch is 20 ft high in the middle, and 120 ft across its major axis. How high above the
water level is the arch, at a point 20 ft from the center (horizontally)? Round off to 2 decimal
places.
Solution: From the illustration, we can see that the unknown variable is the y-coordinate of the
point on the ellipse. To solve this, we must first find the equation of this ellipse from the other
given data. We set the origin as the center of the ellipse, 𝑎 = 60; 𝑏 = 20; 𝑥 = 20
𝑥2 𝑦2
202
𝑦2
+
=
1;
+
= 1;
𝑎2 𝑏 2
602 202
400
𝑦2
400
𝑦2
+
= 1; 1 −
=
3600 400
3600 400
8
𝑦2
=
;
9 400
8
𝑦 2 = ( ) (400) = 355.56
9
𝑦 = √355.56 = 18.86
Answer: The arch is 18.86 feet high above
the water level at a point 20 feet from the
center.
ELABORATE
1. Suppose an ellipse has foci located at (1, 2) and (1, 4) and the point with
coordinates (1, 1) lies on the ellipse. Find the equation of the ellipse.
(𝑦−3)2
(𝑥−1)2
Answer:
+
=1
4
3
2. Find the standard equation of the ellipse whose foci are 𝐹1 (−3, 0) 𝑎𝑛𝑑 𝐹2 (3, 0),
such that for any point on it, the sum of its distances from the foci is 10.
Answer:
𝑥2
25
+
𝑦2
16
=1
3. A tunnel has the shape of a semi-ellipse that is 15 ft high at the center, and 36 ft across at the
base. At most how high should a passing truck be, if it is 12 ft wide, for it to be able to fit through
the tunnel? Round off your answer to two decimal places.
Answer: The passing truck should be 14.14 feet at
most to be able to fit through the tunnel.