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Graphing and Classifying Conics

Lesson 10.6. Graphing and Classifying Conics. Vocabulary. Conic Section: a curve formed by the intersection of a plane and a double napped cone. Examples include parabolas, circles, ellipses, and hyperbolas.

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Graphing and Classifying Conics

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  1. Lesson 10.6 Graphing and Classifying Conics

  2. Vocabulary • Conic Section: a curve formed by the intersection of a plane and a double napped cone. Examples include parabolas, circles, ellipses, and hyperbolas. • General Second-Degree Equation in X and Y: the form Ax2 + Bxy, + Cy2 + Dx + Ey + F = 0. • Discriminant of a Second-Degree Equation: the expression B2 – 4AC for the equation Ax2 +Bxy + Cy2 + Dx + Ey + F = 0. Used to determine which type of conic the equation represents.

  3. 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 = r2 Parabola (y – k)2 = 4p(x – h) (x – h)2 = 4p(y – k) Ellipses and Hyperbola and

  4. Example 1: Writing an Equation of a Translated Parabola • Write an equation of the parabola whose vertex is at (1,-3) and whose focus is at (1,-4). • Write an equation of the parabola whose vertex is at (3,5) and whose focus is at (5,5).

  5. Example 2: Graphing a Translated Circle Graph (x + 1)2 + (y + 2)2 = 4

  6. Example 3: Writing an Equation of a Translated Ellipse • Write an equation of the ellipse with foci at (-1,0) and (3,0) and vertices at (-3,0) and (5,0). • Write an equation of the ellipse with foci at (1,-1) and (-5,-1) and vertices at (4,-1) and (-8,-1).

  7. Example 4: Graphing the Equation of a Translated Hyperbola Graph:

  8. Classifying Conics If the graph of Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is a conic, then the type of conic can be determined as follows: Discriminant Type of Conic B2 – 4AC < 0, B = 0 & A = C Circle B2 – 4AC < 0, either B ≠ 0 or A ≠ 0 Ellipse B2 – 4AC = 0 Parabola B2 – 4AC > 0 Hyperbola

  9. Example 5: Classifying a Conic • x2 + y2 – 2x + 4y – 11 = 0 • y2 – 18x + 84 = 0

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