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 SectionsConic

Section Standard Form of the Equation Example

Parabola

Circle

Ellipse

Hyperbola

PracticeName 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

1. y = x2 + 4 2. y = (x – 3)2 – 2parent function: y = x2; parent function: y = x2; translatedtranslated 4 units up 3 units right and 2 units down

3. y – 1 = x2, or y = x2 + 1 4. y = (x + 5)2 + 6parent function: y = x2; parent function: y = x2;translated 1 unit up translated 5 units left and

6 units up

5. y = x2 – 6x + 1; c = – 2 6. y = x2 + 10x – 7; c = 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

62

102

Solutions

Solutions (continued)

7. y = 2x2 + 8x + 5 8. y = 4x2 – 12x + 3

y = 2(x2 + 4x) + 5; c = 2 y = 4(x2 – 3x) + 3; c = 2

= 22 = 4 =

y = 2(x2 + 4x + 4) + 5 – 2(4) y = 4 x2 – 3x + + 3 – 4

y = 2(x + 2)2 + 5 – 8 y = 4 2 + 3 – 9

y = 2(x + 2)2 – 3 y = 4 2 – 6

42

32–

94

94

94

x –32

x –32

ExampleWrite 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

b2

(y – k)2

a2+ = 1.

(x – (–2))2

42

(y – 4)2

52+ = 1. Substitute –2 for h and 4 for k.

The equation of the ellipse is + = 1.(x + 2)2

16(y – 4)2

25

Example(continued)

(y – 4)2 = (400 – 25(x + 2)2)1

16

Check: Solve the equation for y and graph both equations.

+ = 1.(x + 2)2

16(y – 4)2

25

25(x + 2)2 + 16(y – 4)2 = 400

16(y – 4)2 = 400 – 25(x + 2)2

y – 4 = ± (400 – 25(x + 2)2)1

16

y = 4 ± 400 – 25(x + 2)214

Draw a sketch. The center is the midpoint of the line joining the vertices. Its coordinates are (1, 2).

ExampleWrite an equation of a hyperbola with vertices (–1, 2) and

(3, 2), and foci (–3, 2) and (5, 2).

Find b2 using the Pythagorean Theorem.c2 = a2 + b2

16 = 4 + b2

b2 = 12

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.

Example(continued)

The transverse axis is horizontal, so the equation has the form

(x – h)2

a2

(y – k)2

b2– = 1.

The equation of the hyperbola is(x – 1)2

4(y – 2)2

12– = 1.

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.

ExampleUse 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.

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(x – 40)2

152

y 2

1375– = 1

or (x – 40)2

225y 2

1375– = 1.

ExampleIdentify 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 Group the x- and y- terms.

9(x2 + 2x + ) – 4y2 = 27 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

364y 2

36– = 1 Divide each side by 36.

(x + 1)2

4 – = 1y 2

9 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).