Asked by Anne
Using the matrix [cosx -sinx]
[sinx cosx]
rotate the parabola y=3x^2+2 by 45 degrees. State the equation of the rotated parabola and then state the coordinates of the rotated vertex and the points (-1, 5) and (1, 5)
[sinx cosx]
rotate the parabola y=3x^2+2 by 45 degrees. State the equation of the rotated parabola and then state the coordinates of the rotated vertex and the points (-1, 5) and (1, 5)
Answers
Answered by
Damon
cos45 -sin45
sin45 cos45
is
1/sqrt2 -1/sqrt 2
1/sqrt 2 1/sqrt 2
x' = x/sqrt2 - y/sqrt2
y' = x/sqrt2 + y/sqrt2
so
x/sqrt2 + y/sqrt2 =3(x^2/2-2xy/2+y^2/2) +2
original vertex at (0,2)
x' = 0 - 2/sqrt2
y' = 0 + 2/sqrt2
so
(-sqrt 2 , sqrt 2)
for (-1,5)
x' = -1/sqrt2 -5/sqrt 2 = -6 sqrt2/2 =-3 sqrt 2
y' = -1/sqrt2 +5/sqrt2 = 4/sqrt2
= 2 sqrt 2
etc
sin45 cos45
is
1/sqrt2 -1/sqrt 2
1/sqrt 2 1/sqrt 2
x' = x/sqrt2 - y/sqrt2
y' = x/sqrt2 + y/sqrt2
so
x/sqrt2 + y/sqrt2 =3(x^2/2-2xy/2+y^2/2) +2
original vertex at (0,2)
x' = 0 - 2/sqrt2
y' = 0 + 2/sqrt2
so
(-sqrt 2 , sqrt 2)
for (-1,5)
x' = -1/sqrt2 -5/sqrt 2 = -6 sqrt2/2 =-3 sqrt 2
y' = -1/sqrt2 +5/sqrt2 = 4/sqrt2
= 2 sqrt 2
etc
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