Transcript 10.6 Translating Conic Sections

10.6 – Translating
Conic Sections
Translating Conics means that we move them from the initial
position with an origin at (0, 0) (the parent graph) to a new
position
Equations of Conic Sections
Conic
Section
Parabola
Circle
Ellipse
Hyperbola
Standard Form of the Equation
Example
Practice
Name the parent function for the equations in Exercises 1–4. Describe each
equation as a translation of the parent function.
1. y = x2 + 4
2. y = (x – 3)2 – 2
3. y – 1 = x2
4. y = (x + 5)2 + 6
Rewrite each equation in vertex form. Hint: you may need to complete the
square!
5. y = x2 – 6x + 1
6. y = x2 + 10x – 7
7. y = 2x2 + 8x + 5
8. y = 4x2 – 12x + 3
Solutions
1. y = x2 + 4
parent function: y = x2;
translated 4 units up
2. y = (x – 3)2 – 2
parent function: y = x2; translated
3 units right and 2 units down
3. y – 1 = x2, or y = x2 + 1
parent function: y = x2;
translated 1 unit up
4. y = (x + 5)2 + 6
parent function: y = x2;
translated 5 units left and
6 units up
5. y = x2 – 6x + 1; c =
–6
2
2
6. y = x2 + 10x – 7; c =
10
2
= (–3)2 = 9
= 52 = 25
y = (x2 – 6x + 9) + 1 – 9
y = (x2 + 10x + 25) – 7 – 25
y = (x – 3)2 – 8
y = (x + 5)2 – 32
2
Solutions (continued)
7. y = 2x2 + 8x + 5
y = 2(x2 + 4x) + 5; c =
8. y = 4x2 – 12x + 3
4
2
2
3
2
x2 – 3x + 9 + 3 – 4
9
4
y = 4(x2 – 3x) + 3; c =
= 22 = 4
= 9
y = 2(x2 + 4x + 4) + 5 – 2(4)
y=4
y = 2(x + 2)2 + 5 – 8
y=4 x–
2
3
2
+3–9
y = 2(x + 2)2 – 3
y=4 x– 3
2
2
–6
–2
4
4
Example
Write an equation of an ellipse with center (–2, 4), a vertical major
axis of length 10, and minor axis of length 8.
The length of the major axis is 2a. So 2a = 10 and a = 5.
The length of the minor axis is 2b. So 2b = 8 and b = 4.
Since the center is (–2, 4), h = –2 and k = 4.
The major axis is vertical, so the equation has the form
(x – h)2
(y – k)2
+
= 1.
b2
a2
(x – (–2))2
(y – 4)2
+
= 1.
42
52
Substitute –2 for h and 4 for k.
The equation of the ellipse is
(x + 2)2
(y – 4)2
+ 25 = 1.
16
Example
(continued)
Check: Solve the equation for y and graph both equations.
(x + 2)2
(y – 4)2
+
= 1.
16
25
25(x + 2)2 + 16(y – 4)2 = 400
16(y – 4)2 = 400 – 25(x + 2)2
1
(y – 4)2 = 16 (400 – 25(x + 2)2)
y–4=±
1
2)
(400
–
25(x
+
2)
16
1
y=4± 4
400 – 25(x + 2)2
Example
Write an equation of a hyperbola with vertices (–1, 2) and
(3, 2), and foci (–3, 2) and (5, 2).
Draw a sketch. The center is the
midpoint of the line joining the vertices.
Its coordinates are (1, 2).
The distance between the vertices is 2a,
and the distance between the foci is 2c.
2a = 4, so a = 2; 2c = 8, so c = 4.
Find b2 using the Pythagorean Theorem.
c2 = a2 + b2
16 = 4 + b2
b2 = 12
Example
(continued)
The transverse axis is horizontal, so the equation has the form
(x – h)2
(y – k)2
–
= 1.
a2
b2
The equation of the hyperbola is
(x – 1)2
(y – 2)2
–
= 1.
4
12
Example
Use the information from Example 3. Find the equation of the
hyperbola if the transmitters are 80 mi apart located at (0, 0) and
(80, 0), and all points on the hyperbola are 30 mi closer to one
transmitter than the other.
Since 2c = 80, c = 40. The center of the hyperbola is at (40, 0).
Find a by calculating the difference in the distances
from the vertex at (a + 40, 0) to the two foci.
30 = (a + 40) – (80 – (a + 40))
= 2a
15 = a
(continued)
Find b2.
c2 = a2 + b2
(40)2 = (15)2 + b2
1600 = 225 + b2
b2 = 1375
The equation of the hyperbola is
or
(x – 40)2
y2
–
=1
152
1375
(x – 40)2
y2
–
= 1.
225
1375
Example
Identify the conic section with equation 9x2 – 4y2 + 18x = 27. If it is a
parabola, give the vertex. If it is a circle, give the center and the
radius. If it is an ellipse or a hyperbola, give the center and foci.
Sketch the graph.
Complete the square for the x- and y-terms to write the equation in
standard form.
9x2 – 4y2 + 18x = 27
9x2 + 18x – 4y2 = 27
9(x2 + 2x +
) – 4y2 = 27
Group the x- and y- terms.
Complete the square.
9(x2 + 2x + 1) – 4y2 = 27 + 9(12)
Add (9)(12) to each side.
9(x2 + 2x + 1) – 4y2 = 27 + 9
Simplify
9(x + 1)2 – 4y2 = 36
Write the trinomials as
binomials squared.
Translating Conic Sections
(continued)
9(x + 1)2
4y 2
– 36 = 1
36
(x + 1)2
y2
– 9 =1
4
Divide each side by 36.
Simplify.
The equation represents a hyperbola. The center is (–1, 0). The
transverse axis is horizontal.
Since a2 = 4, a = 2, b2 = 9, so b = 3.
c2 = a2 + b2
=4+9
= 13
c = 13
Translating Conic Sections
(continued)
The distance from the center of the hyperbola to the foci is 13. Since the
hyperbola is centered at (–1, 0), and the transverse axis is horizontal, the
foci are located 13 to the left and right of the center. The foci are at
(–1 + 13, 0) and (–1 – 13, 0).