Table 1

Here is the geometric definition of an ellipse. There are two points called the “foci”: in this case,
(3,0)
and
(3,0)
. A point is on the ellipse if the
sum
of its distances to both foci is a certain constant: in this case, I’ll use
10
. Note that the foci
define
the ellipse, but are not
part of
it.

The point (
x
x
,
y
y) represents any point on the ellipse.
d
1
d1 is its distance from the first focus, and
d
2
d2 to the second. So the ellipse is defined geometrically by the relationship:
d
1
+
d
2
=
10
d1+d2=10.
To calculate
d
1
d1 and
d
2
d2, we use the Pythagorean Theorem as always: drop a straight line down from (
x
x,
y
y) to create the right triangles. Please verify this result for yourself! You should find that
d
1
=
(x+3)2+y2d1=(x+3)2+y2 size 12{ sqrt { \( x+3 \) rSup { size 8{2} } +y rSup { size 8{2} } } } {} and
d
2
=
(x−3)2+y2d2=(x−3)2+y2 size 12{ sqrt { \( x  3 \) rSup { size 8{2} } +y rSup { size 8{2} } } } {}. So the equation becomes:
(x+3)2+y2+
(x−3)2+y2=
10
(x+3)2+y2 size 12{ sqrt { \( x+3 \) rSup { size 8{2} } +y rSup { size 8{2} } } } {}+(x−3)2+y2 size 12{ sqrt { \( x  3 \) rSup { size 8{2} } +y rSup { size 8{2} } } } {}=10. This defines our ellipse
The goal now is to simplify it. We did problems like this earlier in the year (radical equations, the “harder” variety that have two radicals). The way you do it is by isolating the square root, and then squaring both sides. In this case, there are two square roots, so we will need to go through that process twice.
Table 2
(x+3)2+y2=
10
–
(x−3)2+y2(x+3)2+y2 size 12{ sqrt { \( x+3 \) rSup { size 8{2} } +y rSup { size 8{2} } } } {}=10–(x−3)2+y2 size 12{ sqrt { \( x  3 \) rSup { size 8{2} } +y rSup { size 8{2} } } } {} 
Isolate a radical 
(
x
+
3
)
2
+
y
2
=
100
–
20
(x−3)2+y2+
(
x
–
3
)
2
+
y
2
(x+3
)
2
+
y
2
=100–20(x−3)2+y2 size 12{ sqrt { \( x  3 \) rSup { size 8{2} } +y rSup { size 8{2} } } } {}+(x–3
)
2
+
y
2

Square both sides 
(
x
2
+
6
x
+
9
)
+
y
2
=
100
–
20
(x−3)2+y2+
(
x
2
–
6
x
+
9
)
+
y
2
(
x
2
+6x+9)+
y
2
=100–20(x−3)2+y2 size 12{ sqrt { \( x  3 \) rSup { size 8{2} } +y rSup { size 8{2} } } } {}+(
x
2
–6x+9)+
y
2

Multiply out the squares 
12
x
=
100
–
20
(x−3)2+y212x=100–20(x−3)2+y2 size 12{ sqrt { \( x  3 \) rSup { size 8{2} } +y rSup { size 8{2} } } } {} 
Cancel & combine like terms 
(x−3)2+y2=
5
–
35x
(x−3)2+y2 size 12{ sqrt { \( x  3 \) rSup { size 8{2} } +y rSup { size 8{2} } } } {}=5–35 size 12{ { {3} over {5} } } {}x 
Rearrange, divide by 20 
(
x
–
3
)
2
+
y
2
=
25
–
6
x
+
925
x
2
(x–3
)
2
+
y
2
=25–6x+925 size 12{ { {9} over {"25"} } } {}
x
2

Square both sides again 
(
x
2
–
6
x
+
9
)
+
y
2
=
25
–
6
x
+
925x
2
(
x
2
–6x+9)+
y
2
=25–6x+925 size 12{ { {9} over {"25"} } } {}x
2

Multiply out the square 
1625
x
2
+
y
2
=
16
1625 size 12{ { {"16"} over {"25"} } } {}
x
2
+
y
2
=16 
Combine like terms 
x
2
25
+
y
2
16
=
1
x
2
25
+
y
2
16
=
1
size 12{ { {x rSup { size 8{2} } } over {"25"} } + { {y rSup { size 8{2} } } over {"16"} } =1} {}

Divide by 16 
…and we’re done! Now, according to the “machinery” of ellipses, what should that equation look like? Horizontal or vertical? Where should the center be? What are
a
a,
b
b, and
c
c? Does all that match the picture we started with?
"This is the "teacher's guide" book in Kenny Felder's "Advanced Algebra II" series. This text was created with a focus on 'doing' and 'understanding' algebra concepts rather than simply hearing […]"