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**8.6 Translate and Classify Conic Sections**

What is the general 2nd degree equation for any conic? What information can the discriminant tell you about a conic?

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**The equation of any conic can be written in the form-**

Called a general 2nd degree equation

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**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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**Identify the line(s) of symmetry for each conic section in Examples 1 – 4.**

SOLUTION 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.

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**Identify the line(s) of symmetry for the conic section.**

(x – 5)2 64 (y)2 16 7. + = 1 For the ellipse the lines of symmetry are x = 5 and y = 0. ANSWER Identify the line(s) of symmetry for the conic section. 8. (x + 5)2 = 8(y – 2). For parabola the lie of symmetry are x = – 5 ANSWER

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**Identify the line(s) of symmetry for the conic section.**

9. (x – 1)2 49 (y – 2)2 – = 1 121 For horizontal lines of symmetry are x = 1 and y = 2. ANSWER

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Circles Can be multiplied out to look like this….

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Ellipse Can be written like this…..

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Parabola Can be written like this…..

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Hyperbola Can be written like this…..

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**How do you know which conic it is when it’s been multiplied out?**

Pay close attention to whose squared and whose not… Look at the coefficients in front of the squared terms and their signs.

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**Circle Both x and y are squared**

And their coefficients are the same number and sign

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**Ellipse Both x and y are squared**

Their coefficients are different but their signs remain the same.

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Parabola Either x or y is squared but not both

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**Hyperbola Both x and y are squared**

Their coefficients are different and so are their signs.

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Ellipse Parabola Hyperbola Circle You Try!

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**When you want to be sure…**

of a conic equation, then find the type of conic using discriminate information: Ax2 +Bxy +Cy2 +Dx +Ey +F = 0 B2 − 4AC < 0, B = 0 & A = C Circle B2 − 4AC < 0 & either B≠0 or A≠C Ellipse B2 − 4AC = Parabola B2 − 4AC > Hyperbola

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**Classify the Conic 2x2 + y2 −4x − 4 = 0 Ax2 +Bxy +Cy2 +Dx +Ey +F = 0**

B2 − 4AC = 02 − 4(2)(1) = −8 B2 − 4AC < 0, the conic is an ellipse

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**Write the equation in standard form by completing the square**

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**Steps to Complete the Square**

1. Group x’s and y’s. (Boys with the boys and girls with the girls) Send constant numbers to the other side of the equal sign. 2. The coefficient of the x2 and y2 must be 1. If not, factor out. 3. Take the number before the x, divide by 2 and square. Do the same with the number before y. 4. Add these numbers to both sides of the equation. *(Multiply it by the common factor in #2) 5. Factor

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**Graph the Conic 2x2 + y2 −4x − 4 = 0 2x2 −4x + y2 = 4**

V(1±√6), CV(1±√3) Complete the Square

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**Because B2 – 4AC < 0 and A = C, the conic is an circle. **

10. Classify the conic given by x2 + y2 – 2x + 4y + 1 = 0. Then graph the equation. SOLUTION Note that A = 1, B = 0, and C = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(1)(1) = – 4 Because B2 – 4AC < 0 and A = C, the conic is an circle. To graph the circle, first complete the square in both x and y simultaneity . x2 + y2 – 2x + 4y + 1 = 0 x2 – 2x +1+ y2 + 4y + 4 = 4 (x – 1)2 +( y + 2)2 = 4 From the equation, you can see that (h, k) = (– 1, 2), r = 2 Use these facts to draw the circle. ANSWER

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Physical Science In a lab experiment, you record images of a steel ball rolling past a magnet. The equation x2 – 9y2 – 96x + 36y – 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. SOLUTION 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. STEP 1 B2 – 4AC = 02 – 4(16)(– 9) = 576 Because B2 – 4AC > 0, the shape of the path is a hyperbola.

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Write an equation. To write an equation of the hyperbola, complete the square in both x and y simultaneously. STEP 2 16x2 – 9y2 – 96x + 36y – 36 = 0 (16x2 – 96x) – (9y2 – 36y) = 36 16(x2 – 6x + ? ) – 9(y2 – 4y + ? ) = ( ? ) – 9( ? ) 16(x2 – 6x + 9) – 9(y2 – 4y + 4) = (9) – 9(4) 16(x – 3)2 – 9(y – 2)2 = 144 (x – 3)2 9 – (y –2) = 1 STEP 3 Graph the equation. From the equation, the transverse axis is horizontal, (h, k) = (3, 2), a = = 3 and b = = 4 The vertices are at (3 + a, 2), or (6, 2) and (0, 2). See page 530

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**What is the general 2nd degree equation for any conic?**

What information can the discriminant tell you about a conic? B2- 4AC < 0, B = 0, A = C Circle B2- 4AC < 0, B ≠ 0, A ≠ C Ellipse B2- 4AC = 0, Parabola B2- 4AC > 0 Hyperbola

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8.6 Assignment Page 531, odd

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