L9.6 New

Algebra 2:

Algebra 2 Chapter 9 Lesson 6

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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 .

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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.

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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 a 2 = 4 and b 2 = 9 , you know that a = 2 and b = 3 .

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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 ( – 1, 5) and ( – 1, 1) . Because c 2 = a 2 + b 2 = 13 , the foci lie c = 13 3.6 units above and below the center, at ( – 1, 6 .6) and ( – 1, – 0.6) .

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EXAMPLE 2 Graph the equation of a translated hyperbola STEP 3 Draw the hyperbola. Draw a rectangle centered at (–1, 3) that is 2 a = 4 units high and 2 b = 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.

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GUIDED PRACTICE for Examples 1 and 2 Graph ( x + 1) 2 + ( y – 3) 2 = 4. SOLUTION circle with center at ( h , k ) = (– 1, 3) and radius r = 2 1.

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GUIDED PRACTICE for Examples 1 and 2 Graph ( x – 2) 2 = 8 ( y + 3) 2 . SOLUTION parabola with vertex at (2, –3) , focus (2, –1) and directrix y = –5 2.

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GUIDED PRACTICE for Examples 1 and 2 Graph ( x + 3) 2 – ( y – 4) 2 9 = 1 SOLUTION hyperbola with vertices (–4, 4) and (–2, 4), asymtotes y = –2 x – 2 and y = 2 x + 10 3.

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GUIDED PRACTICE for Examples 1 and 2 Graph ( x – 2) 2 16 + ( y – 1) 2 9 = 1 SOLUTION Ellipse with center (2, 1) , vertices (6, 1) and (–2,1) and co–vertices (2, 4) and (2, –2) 4.

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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 = 4 p ( x – h ) where p < 0 .

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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 4 p = – 8 .

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

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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 a 2 + ( y – k ) 2 b 2 = 1

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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 + 2 2 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 .

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EXAMPLE 4 Write an equation of a translated ellipse STEP 4 Find a . For an ellipse , a 2 = b 2 + c 2 = 2 2 + 3 2 = 13, so a = 13 ANSWER The standard form of the equation is ( x – 4) 2 13 + ( y – 2) 2 4 = 1

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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

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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.

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GUIDED PRACTICE for Examples 3, 4 and 5 Write the equation of parabola with vertex at (3, – 1) and focus at (3, 2) . 5. The standard form of the equation is ( x – 3) 2 = 12( y + 1) . ANSWER Write the equation of the hyperbola with vertices at (–7,3) and (–1, 3) and foci at (–9, 3) and (1, 3) . 6. ANSWER The standard form of the equation is ( x + 4) 2 9 – ( y – 3) 2 16 = 1

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GUIDED PRACTICE for Examples 3, 4 and 5 7. ( x – 5) 2 64 + ( y ) 2 16 = 1 Identify the line(s) of symmetry for the conic section. x = 5 and y = 0 . ANSWER

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GUIDED PRACTICE for Examples 3, 4 and 5 8. ( x + 5) 2 = 8( y – 2) . x = –5 ANSWER Identify the line(s) of symmetry for the conic section.

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GUIDED PRACTICE for Examples 3, 4 and 5 9. ( x – 1) 2 49 – ( y – 2) 2 121 = 1 x = 1 and y = 2 . ANSWER Identify the line(s) of symmetry for the conic section.

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

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EXAMPLE 6 Classify a conic 4( x 2 – 2 x + 1 ) + y 2 = 8 + 4( 1 ) 4( x – 1) 2 + y 2 = 12 ( x – 1) 2 3 + y 2 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.

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EXAMPLE 7 Solve a multi-step problem Physical Science In a lab experiment, you record images of a steel ball rolling past a magnet. The equation 16 x 2 – 9 y 2 – 96 x + 36 y – 36 = 0 models the ball’s path. • What is the shape of the path? • Write an equation for the path in standard form. • Graph the equation of the path.

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EXAMPLE 7 Solve a multi-step problem SOLUTION STEP 1 Identify the shape. The equation is a general second-degree equation with A = 16, B = 0, and C = –9 . Find the value of the discriminant. B 2 – 4 AC = 0 2 – 4(16)(–9) = 576 Because B 2 – 4 AC > 0 , the shape of the path is a hyperbola.

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EXAMPLE 7 Solve a multi-step problem STEP 2 Write an equation. To write an equation of the hyperbola, complete the square in both x and y simultaneously. 16 x 2 – 9 y 2 – 96 x + 36 y – 36 = 0 (16 x 2 – 96 x ) – (9 y 2 – 36 y ) = 36 16( x 2 – 6 x + ? ) – 9( y 2 – 4 y + ? ) = 36 + 16( ? ) – 9( ? ) 16( x 2 – 6 x + 9 ) – 9( y 2 – 4 y + 4 ) = 36 + 16( 9 ) – 9( 4 ) 16( x – 3) 2 – 9( y – 2) 2 = 144 ( x – 3) 2 9 – ( y –2) 2 16 = 1

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EXAMPLE 7 Solve a multi-step problem STEP 3 Graph the equation. From the equation, the transverse axis is horizontal, ( h , k ) = (3, 2), a = 9 = 3 and b = 16 . = 4 The vertices are at (3 + a , 2), or (6, 2) and (0, 2) .

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EXAMPLE 7 Solve a multi-step problem STEP 3 Plot the center and vertices. Then draw a rectangle 2 a = 6 units wide and 2 b = 8 units high centered at (3, 2) , draw the asymptotes, and draw the hyperbola. Notice that the path of the ball is modeled by just the right-hand branch of the hyperbola.

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GUIDED PRACTICE for Examples 6 and 7 10. Classify the conic given by x 2 + y 2 – 2 x + 4 y + 1 = 0 . Then graph the equation. ( x – 1) 2 +( y + 2) 2 = 4 ANSWER

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GUIDED PRACTICE for Examples 6 and 7 11. Classify the conic given by 2 x 2 + y 2 – 4 x – 4 = 0 . Then graph the equation. (x – 1) 2 3 y 2 6 + = 1 ANSWER Circle

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GUIDED PRACTICE for Examples 6 and 7 12. Classify the conic given by y 2 – 4 y 2 – 2 x + 6 = 0 . Then graph the equation. ( y – 2) 2 = 2( x –1) ANSWER Parabola

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GUIDED PRACTICE for Examples 6 and 7 13. Classify the conic given by 4 x 2 – y 2 – 16 x – 4 y – 4 = 0 . Then graph the equation. ANSWER ( x –2) 2 4 ( y +2) 2 16 = 1 – Hyperbola

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GUIDED PRACTICE for Examples 6 and 7 14. Astronomy An asteroid’s path is modeled by 4 x 2 + 6.25 y 2 – 12 x – 16 = 0 where x and y are in astronomical units from the sun. Classify the path and write its equation in standard form. Then graph the equation. 4( x – 1.5) 2 4 – y 2 4 = 1 ANSWER Hyperbola