Stitz-Zeager_College_Algebra_e-book

# Substituting x t and 2 2 3 1 y 1 t 2 into 2x 4y 6

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Unformatted text preview: ion. Since we have a sum of squares and the squared terms have unequal coeﬃcients, it’s a good bet we have an ellipse on our hands. (A parabola would have only one squared variable and a circle would have the same coeﬃcient on the squared terms.) We need to complete both squares, and then divide, if necessary, to get the right-hand side equal to 1. x2 + 4y 2 − 2x + 24y + 33 x2 − 2x + 4y 2 + 24y x2 − 2x + 4 y 2 + 6 y 2 − 2x + 1 + 4 y 2 + 6y + 9 x (x − 1)2 + 4(y + 3)2 (x − 1)2 + 4(y + 3)2 4 (x − 1)2 + (y + 3)2 42 (x − 1) (y + 3)2 + 4 1 = = = = = 0 −33 −33 −33 + 1 + 4(9) 4 4 = 4 =1 =1 7.4 Ellipses 425 Now that this equation is in the standard form of Equation 7.4, we see that x − h is x − 1 so h = 1, and y − k is y + 3 so k = −3. Hence, our ellipse is centered at (1, −3). We see that a2 = 4 so a = 2, and b2 = 1 so b = 1. This means we move 2 units left and right from the center and 1 unit up and down from the center to arrive at points on the ellipse. Since we...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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