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.

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.

EXAMPLE 2 Graph the equation of a translated hyperbola

Graph (y – 3)2

4–

(x + 1)2

9= 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.

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 = 13 3.6 units above and below the center, at (– 1, 6.6) and (– 1, – 0.6).

EXAMPLE 2 Graph the equation of a translated hyperbola

STEP 3

Draw the hyperbola. Draw a rectangle centered at (21, 3) that is 2a = 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.

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.

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.

EXAMPLE 3 Write an equation of a translated parabola

The standard form of the equation is (y – 3)2 = – 8(x + 2).

ANSWER

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

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) = 1 + 7 2 + 22 2 )( , = (4, 2)

STEP 3Find 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.

EXAMPLE 4 Write an equation of a translated ellipse

STEP 4

Find a. For an ellipse, a2 = b2 + c2 = 22 + 32 = 13, so a =

13

ANSWER

The standard form of the equation is

(x – 4)2

13 +(y – 2)2

4 = 1

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

EXAMPLE 5 Identify symmetries of conic sections

For the parabola in Example 3, y = 3 is a line of symmetry.

For the ellipse in Example 4, x = 4 and y = 2 are lines of symmetry.

4(x2 – 2x + ? ) + y2 = 8 + 4( ? )

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

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 = 2 3 , and b = 3. Use these facts to draw the ellipse.

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

y2 = 7x – 3

y = + 7x – 3 Equation 1

2x – y = 3

– y = – 2x + 3

y = 2x – 3 Equation 2

EXAMPLE 1 Solve a linear-quadratic system by graphing

STEP 2

Graph the equations y = y = and y = 2x – 3

7x – 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,

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.

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

x2 + (– 3x + 10)2 = 10x2 + 9x2 – 60x + 100 = 10

10x2 – 60x + 90 = 0x2 – 6x + 9 = 0

(x – 3)2 = 0x = 3

Equation 1

Substitute for y.

Expand the power.

Combine like terms.

Divide each side by 10.

Perfect square trinomial

Zero product property

EXAMPLE 2 Solve a linear-quadratic system by substitution

y = – 3(3) + 10 = 1

To find the y-coordinate of the solution, substitute x = 3 in Equation 2.

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).

EXAMPLE 3 Solve a quadratic system by elimination

Solve the system by elimination.9x2 + y2 – 90x + 216 = 0 Equation 1

x2 – y2 – 16 = 0 Equation 2

SOLUTION

9x2 + y2 – 90x + 216 = 0 x2 – y2 – 16 = 0

10x2 – 90x + 200 = 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

Add the equations to eliminate the y2 - term and obtain a quadratic equation in x.

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.

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.

EXAMPLE 4 Solve a real-life quadratic system

x2 – y2 – 16x + 32 = 0 Equation 1

– x2 + y2 – 8y + 8 = 0 Equation 2

SOLUTION

STEP 1Add the equations to eliminate the x2 - and y2 - terms.x2 – y2 – 16x + 32 = 0– x2 + y2 – 8y + 8 = 0

– 16x – 8y + 40 = 0 Add.

y = – 2x + 5 Solve for y.

EXAMPLE 4 Solve a real-life quadratic system

STEP 2Substitute – 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

3x2 – 4x – 7 = 0Substitute for y.

Simplify.

(x + 1)(3x – 7) = 0 Factor.

Zero product propertyx = – 1 or x =73

EXAMPLE 4 Solve a real-life quadratic system

ANSWER

Because the ship is east of the y - axis, it is at ,73

13( ).

STEP 3Substitute for x in y = – 2x + 5 to find the solutions (–1, 7) and ,

73

13( ).