We start by sketching the asymptotes and the vertices

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Unformatted text preview: t most point : ( 7,3) left most point : ( -7,3) top most point : ( 0,5 ) bottom most point : ( 0,1) Here is the sketch of this ellipse. [Return to Problems] (c) Now with this ellipse we’re going to have to be a little careful as it isn’t quite in standard form yet. Here is the standard form for this ellipse. ( x + 1) 2 1 4 © 2007 Paul Dawkins 217 + ( y + 3) = 1 2 http://tutorial.math.lamar.edu/terms.aspx College Algebra Note that in order to get the coefficient of 4 in the numerator of the first term we will need to have a 1 in the denominator. Also, note that we don’t even have a fraction for the y term. This 4 implies that there is in fact a 1 in the denominator. We could put this in if it would be helpful to see what is going on here. ( x + 1) 2 1 4 ( y + 3) + 2 =1 1 So, in this form we can see that the center is ( -1, -3) and that a = 1 and b = 1 . The points for 2 this ellipse are, æ1 ö right most point : ç - , -3 ÷ è2 ø æ3 ö left most point : ç - , -3 ÷ è2 ø top most point :...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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