1. EXAMPLE 1
Graph the equation of a translated circle
Graph (x – 2)2 + (y + 3) 2 = 9.
SOLUTION
STEP 1
Compare the given equation to the standard form of an equation of a circle. You can see that the graph is a circle with center at (h, k) = (2, – 3) and radius r = 9
= 3.

2. EXAMPLE 1
Graph the equation of a translated circle
STEP 2
Plot the center. Then plot several points that are each 3 units from the center:
(2 + 3, – 3) = (5, – 3)
(2 – 3, – 3) = (– 1, – 3)
(2, – 3 + 3) = (2, 0)
(2, – 3 – 3) = (2, – 6)
STEP 3
Draw a circle through the points.

3. EXAMPLE 2
Graph the equation of a translated hyperbola
(y – 3)2 4
(x + 1)2 9
Graph –
= 1
SOLUTION
STEP 1
Compare the given equation to the standard forms of equations of hyperbolas. The equation’s form tells you that the graph is a hyperbola with a vertical transverse axis. The center is at (h, k) = (– 1, 3). Because a2 = 4 and b2 = 9, you know that a = 2 and b = 3.

4. EXAMPLE 2
Graph the equation of a translated hyperbola
STEP 2
Plot the center, vertices, and foci. The vertices lie a = 2 units above and below the center, at (21, 5) and (21, 1). Because c2 = a2 + b2 = 13, the foci lie c = units above and below the center, at (– 1, 6.6) and (– 1, – 0.6).

5. EXAMPLE 2
Graph the equation of a translated hyperbola
STEP 3
Draw the hyperbola. Draw a rectangle centered at (21, 3) that is a = 4 units high and 2b = 6 units wide. Draw the asymptotes through the opposite corners of the rectangle. Then draw the hyperbola passing through the vertices and approaching the asymptotes.

6. EXAMPLE 3
Write an equation of a translated parabola
Write an equation of the parabola whose vertex is at (– 2, 3) and whose focus is at (– 4, 3).
SOLUTION
STEP 1
Determine the form of the equation. Begin by making a rough sketch of the parabola. Because the focus is to the left of the vertex, the parabola opens to the left, and its equation has the form (y – k)2 = 4p(x – h) where p < 0.

7. EXAMPLE 3
Write an equation of a translated parabola
STEP 2
Identify h and k. The vertex is at (– 2, 3), so h = – 2 and k = 3.
STEP 3
Find p. The vertex (– 2, 3) and focus (4, 3) both lie on the line y = 3, so the distance between them is | p | = | – 4 – (– 2) | = 2, and thus p = +2. Because p < 0, it follows that p = – 2, so 4p = – 8.

8. EXAMPLE 3
Write an equation of a translated parabola
ANSWER
The standard form of the equation is (y – 3)2 = – 8(x + 2).

9. EXAMPLE 4
Write an equation of a translated ellipse
Write an equation of the ellipse with foci at (1, 2) and (7, 2) and co-vertices at (4, 0) and (4, 4).
SOLUTION
STEP 1
Determine the form of the equation. First sketch the ellipse. The foci lie on the major axis, so the axis is horizontal. The equation has this form:
(x – h)2 a2 +
(y – k)2 b2
= 1

10. ) ( EXAMPLE 4 Write an equation of a translated ellipse STEP 2
Identify h and k by finding the center, which is halfway between the foci (or the co-vertices)
(h, k) = 2 ) ( ,
= (4, 2)
STEP 3
Find b, the distance between a co-vertex and the center (4, 2), and c, the distance between a focus and the center. Choose the co-vertex (4, 4) and the focus (1, 2): b = | 4 – 2 | = 2 and c = | 1 – 4 | = 3.

11. EXAMPLE 4
Write an equation of a translated ellipse
STEP 4
Find a. For an ellipse, a2 = b2 + c2 = = 13, so a = 13
ANSWER
The standard form of the equation is
(x – 4)2 13 +
(y – 2)2 4
= 1

12. EXAMPLE 5
Identify symmetries of conic sections
Identify the line(s) of symmetry for each conic section in Examples 1 – 4.
SOLUTION
For the circle in Example 1, any line through the center (2, – 3) is a line of symmetry.
For the hyperbola in Example 2 x = – 1 and y = 3 are lines of symmetry

13. EXAMPLE 5
Identify symmetries of conic sections
For the ellipse in Example 4, x = 4 and y = 2 are lines of symmetry.
For the parabola in Example 3, y = 3 is a line of symmetry.

14. EXAMPLE 6
Classify a conic
Classify the conic given by 4x2 + y2 – 8x – 8 = 0. Then graph the equation.
SOLUTION
Note that A = 4, B = 0, and C = 1, so the value of the discriminant is:
B2 – 4AC = 02 – 4(4)(1) = – 16
Because B2– 4AC < 0 and A = C, the conic is an ellipse.
To graph the ellipse, first complete the square in x.
4x2 + y2 – 8x – 8 = 0
(4x2 – 8x) + y2 = 8
4(x2 – 2x) + y2 = 8
4(x2 – 2x + ? ) + y2 = 8 + 4( ? )

15. EXAMPLE 6
Classify a conic
4(x2 – 2x + 1) + y2 = 8 + 4(1)
4(x – 1)2 + y2 = 12
(x – 1)2 3 +
y2 12
= 1
From the equation, you can see that (h, k) = (1, 0), a = 12 = , and b = 3. Use these facts to draw the ellipse.

16. Solve a linear-quadratic system by graphing
EXAMPLE 1
Solve a linear-quadratic system by graphing
Solve the system using a graphing calculator.
y2 – 7x + 3 = 0
Equation 1
2x – y = 3
Equation 2
SOLUTION
STEP 1
Solve each equation for y.
y2 – 7x + 3 = 0
2x – y = 3
y2 = 7x – 3
– y = – 2x + 3
y = x – 3
Equation 1
y = 2x – 3
Equation 2

17. EXAMPLE 1
Solve a linear-quadratic system by graphing
STEP 2
Graph the equations y = y = and y = 2x – 3
7x – 3, –
Use the calculator’s intersect feature to find the coordinates of the intersection points. The graphs of and y = 2x – 3 intersect at (0.75, 21.5). The graphs of and y = 2x – 3 intersect at (4, 5).
y = – 7x – 3,
y = 7x – 3,

18. EXAMPLE 1
Solve a linear-quadratic system by graphing
ANSWER
The solutions are (0.75, – 1.5) and (4, 5). Check the solutions by substituting the coordinates of the points into each of the original equations.

19. Solve a linear-quadratic system by substitution
EXAMPLE 2
Solve a linear-quadratic system by substitution
Solve the system using substitution.
x2 + y2 = 10
Equation 1
y = – 3x + 10
Equation 2
SOLUTION
Substitute –3x + 10 for y in Equation 1 and solve for x.
x2 + y2 = 10
Equation 1
x2 + (– 3x + 10)2 = 10
Substitute for y.
x2 + 9x2 – 60x = 10
Expand the power.
10x2 – 60x + 90 = 0
Combine like terms.
x2 – 6x + 9 = 0
Divide each side by 10.
(x – 3)2 = 0
Perfect square trinomial
x = 3
Zero product property

20. EXAMPLE 2
Solve a linear-quadratic system by substitution
To find the y-coordinate of the solution, substitute x = 3 in Equation 2.
y = – 3(3) + 10 = 1
ANSWER
The solution is (3, 1).
CHECK You can check the solution by graphing the equations in the system. You can see from the graph shown that the line and the circle intersect only at the point (3, 1).

21. Solve a quadratic system by elimination
EXAMPLE 3
Solve a quadratic system by elimination
Solve the system by elimination.
9x2 + y2 – 90x = 0
Equation 1
x2 – y2 – 16 = 0
Equation 2
SOLUTION
Add the equations to eliminate the y2 - term and obtain a quadratic equation in x.
9x2 + y2 – 90x = 0
x2 – y – 16 = 0
10x – 90x = 0
Add.
x2 – 9x + 20 = 0
Divide each side by 10.
(x – 4)(x – 5) = 0
Factor
x = 4 or x = 5
Zero product property

22. EXAMPLE 3
Solve a quadratic system by elimination
When x = 4, y = 0. When x = 5, y = ±3.
ANSWER
The solutions are (4, 0), (5, 3), and (5, 23), as shown.

23. EXAMPLE 4
Solve a real-life quadratic system
Navigation
A ship uses LORAN (long-distance radio navigation) to find its position.Radio signals from stations A and B locate the ship on the blue hyperbola, and signals from stations B and C locate the ship on the red hyperbola.
The equations of the hyperbolas are given below. Find the ship’s position if it is east of the y - axis.

24. Solve a real-life quadratic system
EXAMPLE 4
Solve a real-life quadratic system
x2 – y2 – 16x + 32 = 0
Equation 1
– x2 + y2 – 8y + 8 = 0
Equation 2
SOLUTION
STEP 1
Add the equations to eliminate the x2 - and y2 - terms.
x2 – y2 – 16x = 0
– x2 + y2
– 8y + 8 = 0
– 16x – 8y + 40 = 0
Add.
y = – 2x + 5
Solve for y.

25. Solve a real-life quadratic system
EXAMPLE 4
Solve a real-life quadratic system
STEP 2
Substitute – 2x + 5 for y in Equation 1 and solve for x.
x2 – y2 – 16x + 32 = 0
Equation 1
x2 – (2x + 5)2 – 16x + 32 = 0
Substitute for y.
3x2 – 4x – 7 = 0
Simplify.
(x + 1)(3x – 7) = 0
Factor.
x = – 1 or x = 73
Zero product property

26. ( ). ( ). EXAMPLE 4 Solve a real-life quadratic system STEP 3
Substitute for x in y = – 2x + 5 to find the solutions
(–1, 7) and , 73 13 ( ).
ANSWER
Because the ship is east of the y - axis, it is at , 73 13 ( ).