**Unformatted text preview: **ˆš
is moved to âˆ’ 3 2 2 , 3 2 2 , and (âˆ’3, âˆ’3) is moved to (0, âˆ’3 2). Plotting these in the coordinate
plane along with the lines y = x and y = âˆ’x, we see that the matrix R is rotating these points
counterclockwise by 45â—¦ . y
4
RQ S
2
RS
1 RP
âˆ’4 âˆ’3 âˆ’2 âˆ’1
âˆ’1 1 2 3 Q x âˆ’2
T P âˆ’3
âˆ’4 RT For a generic point P (x, y ) on the hyperbola x2 âˆ’ y 2 = 4, we have
âˆš RP =
= âˆš which means R takes (x, y ) to
2
y = x , we replace x with âˆš 2
2x âˆ’ 2
2x
âˆš
2
2y âˆ’ 2
2
âˆš
2
2
âˆš
2
2x
âˆš
2
2x âˆš
âˆš
2
y, 22 x
2 and y with âˆš âˆ’
âˆ’
+ 2
2
âˆš
2
2
âˆš
2
2y
âˆš
2
2y x
y âˆš +
âˆš 2
2y 2
2x + . To show that this point is on the curve
âˆš 2
2y and simplify. 8.3 Matrix Arithmetic 487 2
2x âˆš 2
2x âˆš âˆ’ 2
2y
âˆš 2
2x âˆš 2
2x 2 2
2y + ? 2
2y âˆš ? 2
2y âˆš âˆ’ 2âˆš
2
xâˆ’ 22 y
2 = âˆš + 2
x y= âˆš âˆš =
2 y2
x2
2âˆ’2
x2 âˆ’ y 2 âˆš 2
2x 2
âˆ’ âˆš âˆš
âˆš 2
2y 2
2x âˆ’ 2
2y ? =2
? =2
=4 Since (x, y ) is on the hyperbola x2 âˆ’ 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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