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C LASSIFYING A C ONIC F ROM ITS E QUATION

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 – 4 AC is called the discriminant

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C LASSIFYING A C ONIC F ROM ITS E QUATION

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  1. 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.

  2. 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.

  3. 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.

  4. 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.

  5. Classify the conic first…then put the equation in standard form. B2 – 4AC 0 – 4(4)(1) –16 < 0 ...so the equation describes an ellipse (since A ≠ C)... Example: Graph the equation 4x2 + y2 – 16x + 2y + 1 = 0 4x2 – 16x +y2 + 2y = –1 Factor out the GCF’s... 4(x2 – 4x ) + (y2 + 2y ) = –1 + 4 +1 + 16 + 1 Complete the squares ... 4(x – 2)2 + (y + 1)2 = 16 4(x – 2)2 + (y + 1)2 = 16 16 16 16 (x – 2)2 + (y +1)2 = 1 4 16 Center (2, –1) a = 2b = 4

  6. 2x2 + 5y2– 4x – 20 = 0 15x2– 5y2 – 30x +20 = 0 Example: Find the points of intersection, if any, of the graphs of 2x2 + 5y2– 4x – 20 = 0 and 3x2– y2 – 6x +4 = 0 5( ) 17x2– 34x = 0 Solve by factoring... 17x(x – 2) = 0 17x = 0 andx – 2 = 0 x = 0andx = 2 Now, find the y-coordinate for each x-coordinate... Solve by linear combinations… 2x2 + 5y2– 4x – 20 = 0 3x2– y2 – 6x + 4 = 0 3x2– y2 – 6x +4 = 0Solved for y2y2 = 3x2–6x +4 when x = 0when x = 2 y2 = 3(0)2– 6(0) + 4 y2 = 3(2)2– 6(2) + 4 y2 = 0 – 0 + 4 y2 = 12 – 12 + 4 y2 = 4 y2 = 4 y= + 2 y = + 2 The points of intersection are (0, 2), (0 –2), (2, 2) and (2, –2)

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