Stitz-Zeager_College_Algebra_e-book

This trend continues and we get c x13 c x12 1 31 1

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Unformatted text preview: is moved to − 3 2 2 , 3 2 2 , and (−3, −3) is moved to (0, −3 2). Plotting these in the coordinate plane along with the lines y = x and y = −x, we see that the matrix R is rotating these points counterclockwise by 45◦ . y 4 RQ S 2 RS 1 RP −4 −3 −2 −1 −1 1 2 3 Q x −2 T P −3 −4 RT For a generic point P (x, y ) on the hyperbola x2 − y 2 = 4, we have √ RP = = √ which means R takes (x, y ) to 2 y = x , we replace x with √ 2 2x − 2 2x √ 2 2y − 2 2 √ 2 2 √ 2 2x √ 2 2x √ √ 2 y, 22 x 2 and y with √ − − + 2 2 √ 2 2 √ 2 2y √ 2 2y x y √ + √ 2 2y 2 2x + . To show that this point is on the curve √ 2 2y and simplify. 8.3 Matrix Arithmetic 487 2 2x √ 2 2x √ − 2 2y √ 2 2x √ 2 2x 2 2 2y + ? 2 2y √ ? 2 2y √ − 2√ 2 x− 22 y 2 = √ + 2 x y= √ √ = 2 y2 x2 2−2 x2 − y 2 √ 2 2x 2 − √ √ √ 2 2y 2 2x − 2 2y ? =2 ? =2 =4 Since (x, y ) is on the hyperbola x2 − y 2 = 4, we know that this last equation is true. Since all of our steps are reversible, this last equation is equivalent to our original equation, which establishes 2 2 the point is,...
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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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