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W RITING AND G RAPHING E QUATIONS OF C ONICS

W RITING AND G RAPHING E QUATIONS OF C ONICS. GRAPHS OF RATIONAL FUNCTIONS. STANDARD FORM OF EQUATIONS OF TRANSLATED CONICS. Horizontal axis. Vertical axis. ( y – k ) 2 = 4 p ( x – h ). ( x – h ) 2 = 4 p ( y – k ). PARABOLA. ( x – h ) 2 ( y – k ) 2.

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W RITING AND G RAPHING E QUATIONS OF C ONICS

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  1. WRITING AND GRAPHING EQUATIONS OF CONICS GRAPHS OF RATIONAL FUNCTIONS STANDARD FORM OF EQUATIONS OF TRANSLATED CONICS Horizontal axis Vertical axis (y – k)2 =4p(x – h) (x – h)2 =4p(y – k) PARABOLA (x – h)2(y – k)2 (x – h)2(y – k)2 + = 1 + = 1 ELLIPSE a2 b2 b2 a2 (x – h)2(y – k)2 (y – k)2(x – h)2 – = 1 – = 1 HYPERBOLA a2 b2 a2 b2 In the following equations the point (h, k) is the vertexof the parabola and the center of the other conics. (x – h)2 + (y – k)2 = r2 CIRCLE

  2. Writing an Equation of a Translated Parabola (–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) Write an equation of the parabola whose vertex is at (–2, 1) and whose focus is at (–3, 1). SOLUTION Find h and k: The vertex is at(–2, 1), so h = –2 and k = 1.

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

  4. Graphing the Equation of a Translated Circle Graph (x – 3)2+ (y + 2)2 = 16. SOLUTION Compare the given equation to the standard form of the equation of a circle: (3, – 2) (x – h)2+ (y – k)2= r2 You can see that the graph will be a circle with center at (h, k) = (3, – 2).

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

  6. Writing an Equation of a Translated Ellipse (3, 6) (3, 5) The ellipse has a vertical major axis, so its equation is of the form: (x – h)2(y – k)2 + = 1 b2 a2 Find the center: The center is halfwaybetween the vertices. (3, –2) (3, –1) (3 + 3) 6 + ( –2) (h, k) = , = (3, 2) 2 2 Write an equation of the ellipse with foci at (3,5) and (3, –1) and vertices at (3,6) and (3, –2). SOLUTION Plot the given points and make a rough sketch.

  7. Writing an Equation of a Translated Ellipse (3, 6) (3, 5) a = (3 – 3)2 + (6 – 2)2= 0 + 42= 4 (3, –2) (3, –1) c= (3 – 3)2 + (5 – 2)2= 0 + 32= 3 Write an equation of the ellipse with foci at (3,5) and (3, –1) and vertices at (3,6) and (3, –2). SOLUTION Find a: The value of a is the distancebetween the vertex and the center. Find c: The value of c is the distancebetween the focus and the center.

  8. Writing an Equation of a Translated Ellipse (3, 6) (3, 5) b2 = 42–32 b2 = 7 b = 7 (3, –2) (3, –1) (x – 3)2(y – 2)2 The standard form is + = 1 7 16 Write an equation of the ellipse with foci at (3,5) and (3, –1) and vertices at (3,6) and (3, –2). SOLUTION Find b: Substitute the values of aandc into the equation b2 = a2–c2.

  9. Graphing the Equation of a Translated Hyperbola (x + 1)2 Graph (y + 1)2– = 1. 4 (–1, 0) (–1, –1) (–1, –2) SOLUTION The y2-term is positive, so thetransverse axis is vertical. Sincea2 = 1 and b2 = 4, you know thata = 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.

  10. Graphing the Equation of a Translated Hyperbola (x + 1)2 Graph (y + 1)2– = 1. 4 (–1, 0) (–1, –1) (–1, –2) SOLUTION The y2-term is positive, so thetransverse axis is vertical. Sincea2 = 1 and b2 = 4, you know thata = 1 and b = 2. Draw the asymptotes through the corners of the rectangle. Draw the hyperbola so that it passes through the vertices and approaches the asymptotes.

  11. CLASSIFYING A CONIC FROM ITS EQUATION The equation of any conic can be written in the form Ax2+ Bxy + Cy2+ Dx + Ey+ F= 0 which is called a general second-degree equation in x and y. The expression B2 – 4AC is called the discriminant of the equation and can be used to determine which typeof conic the equation represents.

  12. CONIC TYPES CONCEPT SUMMARY (B 2 – 4AC) DISCRIMINANT CLASSIFYING ACONICFROM ITS EQUATION The type of conic can be determined as follows: TYPE OF CONIC < 0, B = 0, and A = C Circle < 0, and either B 0, or A C Ellipse = 0 Parabola > 0 Hyperbola If B = 0, each axis is horizontal or vertical. If B 0, the axes are neither horizontal nor vertical.

  13. Classifying a Conic Classify the conic 2x2 + y2 – 4x – 4 = 0. Help SOLUTION Since A = 2, B = 0, and C = 1, the value of the discriminant is: B2 – 4AC = 02– 4(2)(1) = –8 Because B2– 4AC < 0 and AC, the graph is an ellipse.

  14. Classifying a Conic Classify the conic 4x2 – 9y2 + 32x – 144y – 548 = 0. Help SOLUTION Since A = 4, B = 0, and C = –9, the value of the discriminant is: B2– 4AC = 02– 4(4)(–9) = 144 Because B2– 4AC> 0, the graph is a hyperbola.

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