Stitz-Zeager_College_Algebra_e-book

We begin our study of systems of equations by

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Unformatted text preview: that fact (0, b) is on the ellipse, along with the fact that d = 2a to get distance from (−c, 0) to (0, b) + distance from (c, 0) to (0, b) (0 − (−c))2 + (b − 0)2 + (0 − c)2 + (b − 0)2 √ √ b2 + c2 + b2 + c2 √ 2 b2 + c2 √ b2 + c2 = = = = = 2a 2a 2a 2a a From this, we get a2 = b2 + c2 , or b2 = a2 − c2 , which will prove useful later. Now consider a point (x, y ) on the ellipse. Applying Deﬁnition 7.4, we get distance from (−c, 0) to (x, y ) + distance from (c, 0) to (x, y ) = 2a (x − (−c))2 + (y − 0)2 + (x − c)2 + (y − 0)2 = 2a (x + c)2 + y 2 + (x − c)2 + y 2 = 2a In order to make sense of this situation, we need to do some rearranging, squaring, and more rearranging.1 (x + c)2 + y 2 + (x − c)2 + y 2 = 2a (x + c)2 + y 2 = 2a − (x + c)2 + y 2 4a 4a a a 2 (x + c)2 + y 2 (x − c)2 + y 2 (x − c)2 + y 2 (x − c)2 + y 2 (x − c)2 + y 2 2 a2 (x − c)2 + y 2 a2 x2 − 2a2 cx + a2 c2 + a2 y 2 a2 x2 − c2 x2 + a2 y 2 a2 − c2 x2 + a2 y 2 = = = = = = = = = = (x − c)2 + y 2 2a − (x − c)2 + y 2 2 4a2 − 4a (x − c)2 + y 2 + (x − c)2 + y 2 4a2 + (x...
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