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 Jan 4 at 9:51
Jimmy SabaterJimmy Sabater
1,980219
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 Jan 4 at 9:51
Jimmy SabaterJimmy Sabater
1,980219
Plot the two graphs and see.
– amd
Jan 4 at 20:09
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 Jan 4 at 9:51
Jimmy SabaterJimmy Sabater
1,980219
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 Jan 4 at 9:51
Jimmy SabaterJimmy Sabater
1,980219
asked Jan 4 at 9:51
Jimmy SabaterJimmy Sabater
1,980219
asked Jan 4 at 9:51
Jimmy SabaterJimmy Sabater
1,980219
asked Jan 4 at 9:51
Jimmy SabaterJimmy Sabater
1,980219
asked Jan 4 at 9:51
Jimmy SabaterJimmy Sabater
1,980219
1,980219
Plot the two graphs and see.
– amd
Jan 4 at 20:09
add a comment |
Plot the two graphs and see.
– amd
Jan 4 at 20:09
Plot the two graphs and see.
– amd
Jan 4 at 20:09
Plot the two graphs and see.
– amd
Jan 4 at 20:09
add a comment |
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Plot the two graphs and see.
– amd
Jan 4 at 20:09