Understanding rotations of graphs about the origin
Suppose the hyperbola $x^2 - y^2 = 1$ is rotated anti clockwise about
the origin thru $pi/4$ . ${bf find ; its ; new ; equation}$.
I know that if we rotate the usual coordinate xy-plane about an angle $alpha$ anti clockwise, then the new coordinates are given by
$$ begin{bmatrix} x' \ y' end{bmatrix} = begin{bmatrix} cos alpha & sin alpha \ - sin alpha & cos alpha end{bmatrix}begin{bmatrix} x \ y end{bmatrix} $$
My question is: Rotating the hyperbola thru 45 degrees anti clockwise is it the same as to rotate the axis the same amount? IN such case then we have
$$ begin{bmatrix} x' \ y' end{bmatrix} = frac{1}{sqrt{2}}begin{bmatrix} 1 & 1 \ - 1 & 1 end{bmatrix}begin{bmatrix} x \ y end{bmatrix} = frac{1}{sqrt{2}} begin{bmatrix} x+y \ y-x end{bmatrix} $$
Now, we see that
$$ x' + y' = sqrt{2} y $$ and $$x' - y' = sqrt{2} x $$
Thus,
$$ x^2 - y^2 = frac{1}{2} ((x'+y')^2 - (x'-y')^2) = 1 $$
which implies
$$ ( 2y ') ( 2x' ) = 2 implies boxed{ x' y' = 1/2 }$$
is this correct?
calculus linear-algebra
asked 13 hours ago
Jimmy Sabater
1,937219
add a comment |
Suppose the hyperbola $x^2 - y^2 = 1$ is rotated anti clockwise about
the origin thru $pi/4$ . ${bf find ; its ; new ; equation}$.
I know that if we rotate the usual coordinate xy-plane about an angle $alpha$ anti clockwise, then the new coordinates are given by
$$ begin{bmatrix} x' \ y' end{bmatrix} = begin{bmatrix} cos alpha & sin alpha \ - sin alpha & cos alpha end{bmatrix}begin{bmatrix} x \ y end{bmatrix} $$
My question is: Rotating the hyperbola thru 45 degrees anti clockwise is it the same as to rotate the axis the same amount? IN such case then we have
$$ begin{bmatrix} x' \ y' end{bmatrix} = frac{1}{sqrt{2}}begin{bmatrix} 1 & 1 \ - 1 & 1 end{bmatrix}begin{bmatrix} x \ y end{bmatrix} = frac{1}{sqrt{2}} begin{bmatrix} x+y \ y-x end{bmatrix} $$
Now, we see that
$$ x' + y' = sqrt{2} y $$ and $$x' - y' = sqrt{2} x $$
Thus,
$$ x^2 - y^2 = frac{1}{2} ((x'+y')^2 - (x'-y')^2) = 1 $$
which implies
$$ ( 2y ') ( 2x' ) = 2 implies boxed{ x' y' = 1/2 }$$
is this correct?
calculus linear-algebra
asked 13 hours ago
Jimmy Sabater
1,937219
Plot the two graphs and see.
– amd
3 hours ago
add a comment |
Suppose the hyperbola $x^2 - y^2 = 1$ is rotated anti clockwise about
the origin thru $pi/4$ . ${bf find ; its ; new ; equation}$.
I know that if we rotate the usual coordinate xy-plane about an angle $alpha$ anti clockwise, then the new coordinates are given by
$$ begin{bmatrix} x' \ y' end{bmatrix} = begin{bmatrix} cos alpha & sin alpha \ - sin alpha & cos alpha end{bmatrix}begin{bmatrix} x \ y end{bmatrix} $$
My question is: Rotating the hyperbola thru 45 degrees anti clockwise is it the same as to rotate the axis the same amount? IN such case then we have
$$ begin{bmatrix} x' \ y' end{bmatrix} = frac{1}{sqrt{2}}begin{bmatrix} 1 & 1 \ - 1 & 1 end{bmatrix}begin{bmatrix} x \ y end{bmatrix} = frac{1}{sqrt{2}} begin{bmatrix} x+y \ y-x end{bmatrix} $$
Now, we see that
$$ x' + y' = sqrt{2} y $$ and $$x' - y' = sqrt{2} x $$
Thus,
$$ x^2 - y^2 = frac{1}{2} ((x'+y')^2 - (x'-y')^2) = 1 $$
which implies
$$ ( 2y ') ( 2x' ) = 2 implies boxed{ x' y' = 1/2 }$$
is this correct?
calculus linear-algebra
asked 13 hours ago
Jimmy Sabater
1,937219
Suppose the hyperbola $x^2 - y^2 = 1$ is rotated anti clockwise about
the origin thru $pi/4$ . ${bf find ; its ; new ; equation}$.
I know that if we rotate the usual coordinate xy-plane about an angle $alpha$ anti clockwise, then the new coordinates are given by
$$ begin{bmatrix} x' \ y' end{bmatrix} = begin{bmatrix} cos alpha & sin alpha \ - sin alpha & cos alpha end{bmatrix}begin{bmatrix} x \ y end{bmatrix} $$
My question is: Rotating the hyperbola thru 45 degrees anti clockwise is it the same as to rotate the axis the same amount? IN such case then we have
$$ begin{bmatrix} x' \ y' end{bmatrix} = frac{1}{sqrt{2}}begin{bmatrix} 1 & 1 \ - 1 & 1 end{bmatrix}begin{bmatrix} x \ y end{bmatrix} = frac{1}{sqrt{2}} begin{bmatrix} x+y \ y-x end{bmatrix} $$
Now, we see that
$$ x' + y' = sqrt{2} y $$ and $$x' - y' = sqrt{2} x $$
Thus,
$$ x^2 - y^2 = frac{1}{2} ((x'+y')^2 - (x'-y')^2) = 1 $$
which implies
$$ ( 2y ') ( 2x' ) = 2 implies boxed{ x' y' = 1/2 }$$
is this correct?
calculus linear-algebra
calculus linear-algebra
asked 13 hours ago
Jimmy Sabater
1,937219
asked 13 hours ago
Jimmy Sabater
1,937219
asked 13 hours ago
Jimmy Sabater
1,937219
asked 13 hours ago
Jimmy Sabater
1,937219
asked 13 hours ago
Jimmy Sabater
1,937219
1,937219
Plot the two graphs and see.
– amd
3 hours ago
add a comment |
Plot the two graphs and see.
– amd
3 hours ago
Plot the two graphs and see.
– amd
3 hours ago
Plot the two graphs and see.
– amd
3 hours ago
add a comment |
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Plot the two graphs and see.
– amd
3 hours ago