MCR3U - Santowski
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ex 1. Given the circle x 2 + y 2 = 4 we will apply the following transformation T(x,y) => (2x,y) which is interpreted as a horizontal stretch by a factor of 2.
ex 2. Given the circle x 2 + y 2 = 4, apply the transformation T(x,y) => (x,3y) which is interpreted as a vertical stretch by a factor of 3
From these two transformations, we can see that we have formed a new shape, which is called an ellipse
2
3
An ellipse is defined as the set of points such that the sum of the distances from any point on the ellipse to two stationary points (called the foci) is a constant
We will explore the ellipse from a locus definition in two ways
ex 3. Using grid paper with 2 sets of concentric circles, we can define the two circle centers as fixed points and then label all other points (P), that meet the requirement that the sum of the distances from the point (P) on the ellipse to the two fixed centers (which we will call foci) will be a constant i.e. PF
1
PF
2
= constant. We will work with the example that PF
1
+ PF
2
= 10 units.
+
ex 4. Using the GSP program, we will geometrically construct a set of points that satisfy the condition that PF following link
1
+ PF
2
= constant by following the
4
5
We will now tie in our knowledge of algebra to come up with an algebraic description of the ellipse by making use of the relationship that PF
1
+ PF
2
= constant
ex 5. Find the equation of the ellipse whose foci are at (+3,0) and the constant (which is called the sum of the focal radi) is 10. Then sketch the ellipse by finding the x and y intercepts.
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Since we are dealing with distances, we set up our equation using the general point P(x,y), F
1 at (-3,0) and F
2 at (3,0) and the algebra follows on the next slide |PF
1
| + |PF
2
| = 10
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PF
1
PF
2
x
3
2
10
y
2
x
3
2 y
2
10
x x
2
x
3
2
3
2 y
2
6 x
9
y
2
2
10
100
20
x
x
3
2
3
2 y
2 y
2
100
20
x
3
2 y
2
y
2
2
x
3
2
x
2
y
2
6 x
9
y
2
20
x
3
2 y
2
100
12 x
5
x
3
2
25
x
2
y
2
6 x
9
2
y
2
25
3 x
2
625
150 x
9 x
2
25 x
2
9 x
2
150 x
150 x
25 y
2
625
225
16 x
2
25 y
2
400
16 x 2
400
25 y 2
400
x 2
25
y 2
16
x
5
2
y
4
2
1
8
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The equation of the ellipse is (x/5) 2 + (y/4) 2 = 1 OR 16x 2 + 25y 2 = 400
The x-intercepts occur at (+5,0) and the y-intercepts occur at (0,+4)
The domain is {x E R | -5 < x < 5} and the range is {y E R | -4 < y < 4}
NOTE that this is NOT a function, but rather a relation
NOTE the relationship between the equation and the intercepts, domain and range
so to generalize, if the ellipse has the
2
2
, then the x-intercepts occur at
(+a,0), the y-intercepts at (0,+b) and the domain is between –a and +a and the range is between –b and +b OR we can rewrite the equation in the form of (bx) 2 + (ay) 2 = (ab) 2
Note that if a > b, then the ellipse is longer along the x-axis than along the y-axis
so if a < b, then the ellipse would be longer along the y-axis
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The longer of the two axis is called the major axis and lies between the 2 x-intercepts
(if a > b ). Its length is 2 a
The shorter of the two axis is called the minor axis and lies between the 2 y-intercepts
(if a > b ). Its length is 2 b
The two end points of the major axis (in this case the x-intercepts) are called vertices
(at (+a,0))
The two foci lie on the major axis
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Determine the equation of the ellipse and then sketch it, labelling the key features, if the foci are at (+4, 0) and the sum of the focal radii is 12 units
(i.e.
the fancy name for the constant distance sum PF
1
+ PF
2
)
The equation you generate should be x 2 /36 + y 2 /20 = 1
12
AW, p470, Q8bc, 9bc
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http://www.analyzemath.com/EllipseEq/EllipseE q.html
- an interactive applet fom AnalyzeMath http://home.alltel.net/okrebs/page62.html
-
Examples and explanations from OJK's
Precalculus Study Page http://tutorial.math.lamar.edu/AllBrowsers/1314/
Ellipses.asp
- Ellipses from Paul Dawkins at
Lamar University http://www.webmath.com/ellipse1.html
- Graphs of ellipses from WebMath.com
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