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W RITING AND G RAPHING E QUATIONS OF C ONICS GRAPHS OF RATIONAL FUNCTIONS STANDARD FORM OF EQUATIONS OF TRANSLATED CONICS In the following equations the point (h, k) is the vertex of the parabola and the center of the other conics. CIRCLE (x – h) 2 + (y – k) 2 = r 2 Horizontal axisVertical axis PARABOLA (y – k) 2 = 4 p (x – h)(x – h) 2 = 4 p (y – k) HYPERBOLA (x – h) 2 (y – k) 2 – = 1 b 2b 2 a 2a 2 (y – k) 2 (x – h) 2 – = 1 b 2b 2 a 2a 2 ELLIPSE (x – h) 2 (y – k) 2 + = 1 a 2a 2 b 2b 2 (x – h) 2 (y – k) 2 + = 1 a 2a 2 b 2b 2

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Writing an Equation of a Translated Parabola Write an equation of the parabola whose vertex is at (–2, 1) and whose focus is at (–3, 1). S OLUTION (–2, 1) Choose form: Begin by sketching the parabola. Because the parabola opens to the left, it has the form where p < 0. (y – k) 2 = 4p(x – h) Find h and k: The vertex is at (–2, 1), so h = – 2 and k = 1.

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Writing an Equation of a Translated Parabola Write an equation of the parabola whose vertex is at (–2, 1) and whose focus is at (–3, 1). S OLUTION (–3, 1) (–2, 1) The standard form of the equation is (y – 1) 2 = – 4(x + 2). Find p: The distance between the vertex (–2, 1), and the focus (–3, 1) is p = (–3 – (–2)) 2 + (1 – 1) 2 = 1 so p = 1 or p = – 1. Since p < 0, p = – 1.

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Graphing the Equation of a Translated Circle Graph (x – 3) 2 + (y + 2) 2 = 16. S OLUTION Compare the given equation to the standard form of the equation of a circle: (x – h) 2 + (y – k) 2 = r 2 You can see that the graph will be a circle with center at (h, k) = (3, – 2). (3, – 2)

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The radius is r = 4 Graphing the Equation of a Translated Circle ( 3 + 4, – 2 + 0) = (7, – 2) ( 3 + 0, – 2 + 4 ) = (3, 2) ( 3 – 4, – 2 + 0) = (– 1, – 2) ( 3 + 0, – 2 – 4 ) = (3, – 6) Draw a circle through the points. Graph (x – 3) 2 + (y + 2) 2 = 16. S OLUTION (– 1, – 2) (3, – 6) (3, 2) (3, – 2) r Plot several points that are each 4 units from the center: (7, – 2)

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Writing an Equation of a Translated Ellipse Write an equation of the ellipse with foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2). S OLUTION Plot the given points and make a rough sketch. (x – h) 2 (y – k) 2 + = 1 a 2a 2 b 2b 2 The ellipse has a vertical major axis, so its equation is of the form: (3, 5) (3, –1) (3, 6) (3, –2) Find the center: The center is halfway between the vertices. (3 + 3) 6 + ( –2) 2 (h, k) =, = (3, 2) 2

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S OLUTION (3, 5) (3, –1) (3, 6) (3, –2) Writing an Equation of a Translated Ellipse Find a: The value of a is the distance between the vertex and the center. Find c: The value of c is the distance between the focus and the center. a = (3 – 3) 2 + (6 – 2) 2 = 0 + 4 2 = 4 c = (3 – 3) 2 + (5 – 2) 2 = 0 + 3 2 = 3 Write an equation of the ellipse with foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).

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S OLUTION (3, 5) (3, –1) (3, 6) (3, –2) Writing an Equation of a Translated Ellipse Find b: Substitute the values of a and c into the equation b 2 = a 2 – c 2. b 2 = 4 2 – 3 2 b 2 = 7 b = 7 16 7 + = 1 The standard form is (x – 3 ) 2 (y – 2 ) 2 Write an equation of the ellipse with foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).

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Graphing the Equation of a Translated Hyperbola Graph (y + 1) 2 – = 1. (x + 1) 2 4 S OLUTION The y 2 -term is positive, so the transverse axis is vertical. Since a 2 = 1 and b 2 = 4, you know that a = 1 and b = 2. Plot the center at (h, k) = (–1, –1). Plot the vertices 1 unit above and below the center at (–1, 0) and (–1, –2). Draw a rectangle that is centered at (–1, –1) and is 2a = 2 units high and 2b = 4 units wide. (–1, –2) (–1, 0) (–1, –1)

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Graphing the Equation of a Translated Hyperbola S OLUTION Draw the asymptotes through the corners of the rectangle. Draw the hyperbola so that it passes through the vertices and approaches the asymptotes. (–1, –2) (–1, 0) (–1, –1) Graph (y + 1) 2 – = 1. (x + 1) 2 4 The y 2 -term is positive, so the transverse axis is vertical. Since a 2 = 1 and b 2 = 4, you know that a = 1 and b = 2.

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C LASSIFYING A C ONIC F ROM ITS E QUATION The equation of any conic can be written in the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 which is called a general second-degree equation in x and y. The expression B 2 – 4AC is called the discriminant of the equation and can be used to determine which type of conic the equation represents.

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CONIC TYPES C LASSIFYING A C ONIC F ROM ITS E QUATION C ONCEPT S UMMARY The type of conic can be determined as follows: TYPE OF CONIC DISCRIMINANT (B 2 – 4AC) < 0,B = 0,and A = C < 0, and either B 0, or A C = 0 > 0 Circle Ellipse Parabola Hyperbola If B = 0, each axis is horizontal or vertical. If B 0, the axes are neither horizontal nor vertical.

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Classifying a Conic Classify the conic 2 x 2 + y 2 – 4 x – 4 = 0. S OLUTION Since A = 2, B = 0, and C = 1, the value of the discriminant is: B 2 – 4 AC = 0 2 – 4 (2) (1) = – 8 Because B 2 – 4 AC < 0 and A C, the graph is an ellipse. Help

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Classifying a Conic Classify the conic 4 x 2 – 9 y 2 + 32 x – 144 y – 5 48 = 0. S OLUTION Since A = 4, B = 0, and C = –9, the value of the discriminant is: B 2 – 4 AC = 0 2 – 4 (4) (–9) = 144 Because B 2 – 4 AC > 0, the graph is a hyperbola. Help

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CONIC TYPES C LASSIFYING A C ONIC F ROM ITS E QUATION C ONCEPT S UMMARY The type of conic can be determined as follows: TYPE OF CONIC DISCRIMINANT (B 2 – 4AC) < 0,B = 0,and A = C < 0, and either B 0, or A C = 0 > 0 Circle Ellipse Parabola Hyperbola If B = 0, each axis is horizontal or vertical. If B 0, the axes are neither horizontal nor vertical. Back

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CONIC TYPES C LASSIFYING A C ONIC F ROM ITS E QUATION C ONCEPT S UMMARY The type of conic can be determined as follows: TYPE OF CONIC DISCRIMINANT (B 2 – 4AC) < 0,B = 0,and A = C < 0, and either B 0, or A C = 0 > 0 Circle Ellipse Parabola Hyperbola If B = 0, each axis is horizontal or vertical. If B 0, the axes are neither horizontal nor vertical. Back

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