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Writing and Graphing Equations of Conics

This article explains how to write and graph equations of conics, such as parabolas, ellipses, and hyperbolas. It also covers the standard form of equations and translated conics. Examples and step-by-step solutions are provided.

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Writing and Graphing Equations of Conics

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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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