Simplifying an ellipse by change of bases
up vote
0
down vote
favorite
ellipse = $2x^2 - 4xy + 5y^2 = 1$
I found the transition matrix, P, for the given bases:
B' = {(2/$sqrt5$, 1/$sqrt5$),(-1/$sqrt5$, 2/$sqrt5$)
B = {(1,0), (0,1)}
P = $begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix}$
2) Convert a coordinate matrix $begin{bmatrix}x'\y'end{bmatrix}$ relative to the basis $B'$ into the coordinate matrix $begin{bmatrix}x\yend{bmatrix}$ relative to the basis $B$. Plug both $x,y$ found in 2) to the original equation and find a simplified equation in terms of the variables $x', y'$
Here is where I am confused by what the question is asking. I chose values for x' and y'. $begin{bmatrix}x'=1\y'=1end{bmatrix}$ and then plug it in and get x, y values.
$begin{bmatrix}x\yend{bmatrix}$ = $begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix}$ $begin{bmatrix}1\1end{bmatrix}$ = $begin{bmatrix}
sqrt2
\
sqrt1.8
end{bmatrix}$
Here is where I am stuck. How am i supposed to plug the x and y values into the eclipse equation to make it a "simplified" equation?
linear-algebra
asked Nov 27 at 0:12
Evan Kim
758
add a comment |
up vote
0
down vote
favorite
ellipse = $2x^2 - 4xy + 5y^2 = 1$
I found the transition matrix, P, for the given bases:
B' = {(2/$sqrt5$, 1/$sqrt5$),(-1/$sqrt5$, 2/$sqrt5$)
B = {(1,0), (0,1)}
P = $begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix}$
2) Convert a coordinate matrix $begin{bmatrix}x'\y'end{bmatrix}$ relative to the basis $B'$ into the coordinate matrix $begin{bmatrix}x\yend{bmatrix}$ relative to the basis $B$. Plug both $x,y$ found in 2) to the original equation and find a simplified equation in terms of the variables $x', y'$
Here is where I am confused by what the question is asking. I chose values for x' and y'. $begin{bmatrix}x'=1\y'=1end{bmatrix}$ and then plug it in and get x, y values.
$begin{bmatrix}x\yend{bmatrix}$ = $begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix}$ $begin{bmatrix}1\1end{bmatrix}$ = $begin{bmatrix}
sqrt2
\
sqrt1.8
end{bmatrix}$
Here is where I am stuck. How am i supposed to plug the x and y values into the eclipse equation to make it a "simplified" equation?
linear-algebra
asked Nov 27 at 0:12
Evan Kim
758
Small suggestion: the thing you're simplifying is an ellipse; an eclipse is an occultation of one celestial body by another.
– John Hughes
Nov 27 at 0:14
thank you.......
– Evan Kim
Nov 27 at 0:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
ellipse = $2x^2 - 4xy + 5y^2 = 1$
I found the transition matrix, P, for the given bases:
B' = {(2/$sqrt5$, 1/$sqrt5$),(-1/$sqrt5$, 2/$sqrt5$)
B = {(1,0), (0,1)}
P = $begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix}$
2) Convert a coordinate matrix $begin{bmatrix}x'\y'end{bmatrix}$ relative to the basis $B'$ into the coordinate matrix $begin{bmatrix}x\yend{bmatrix}$ relative to the basis $B$. Plug both $x,y$ found in 2) to the original equation and find a simplified equation in terms of the variables $x', y'$
Here is where I am confused by what the question is asking. I chose values for x' and y'. $begin{bmatrix}x'=1\y'=1end{bmatrix}$ and then plug it in and get x, y values.
$begin{bmatrix}x\yend{bmatrix}$ = $begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix}$ $begin{bmatrix}1\1end{bmatrix}$ = $begin{bmatrix}
sqrt2
\
sqrt1.8
end{bmatrix}$
Here is where I am stuck. How am i supposed to plug the x and y values into the eclipse equation to make it a "simplified" equation?
linear-algebra
asked Nov 27 at 0:12
Evan Kim
758
ellipse = $2x^2 - 4xy + 5y^2 = 1$
I found the transition matrix, P, for the given bases:
B' = {(2/$sqrt5$, 1/$sqrt5$),(-1/$sqrt5$, 2/$sqrt5$)
B = {(1,0), (0,1)}
P = $begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix}$
2) Convert a coordinate matrix $begin{bmatrix}x'\y'end{bmatrix}$ relative to the basis $B'$ into the coordinate matrix $begin{bmatrix}x\yend{bmatrix}$ relative to the basis $B$. Plug both $x,y$ found in 2) to the original equation and find a simplified equation in terms of the variables $x', y'$
Here is where I am confused by what the question is asking. I chose values for x' and y'. $begin{bmatrix}x'=1\y'=1end{bmatrix}$ and then plug it in and get x, y values.
$begin{bmatrix}x\yend{bmatrix}$ = $begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix}$ $begin{bmatrix}1\1end{bmatrix}$ = $begin{bmatrix}
sqrt2
\
sqrt1.8
end{bmatrix}$
Here is where I am stuck. How am i supposed to plug the x and y values into the eclipse equation to make it a "simplified" equation?
linear-algebra
linear-algebra
asked Nov 27 at 0:12
Evan Kim
758
asked Nov 27 at 0:12
Evan Kim
758
edited Nov 27 at 0:15
asked Nov 27 at 0:12
Evan Kim
758
asked Nov 27 at 0:12
Evan Kim
758
asked Nov 27 at 0:12
Evan Kim
758
758
Small suggestion: the thing you're simplifying is an ellipse; an eclipse is an occultation of one celestial body by another.
– John Hughes
Nov 27 at 0:14
thank you.......
– Evan Kim
Nov 27 at 0:15
add a comment |
Small suggestion: the thing you're simplifying is an ellipse; an eclipse is an occultation of one celestial body by another.
– John Hughes
Nov 27 at 0:14
thank you.......
– Evan Kim
Nov 27 at 0:15
Small suggestion: the thing you're simplifying is an ellipse; an eclipse is an occultation of one celestial body by another.
– John Hughes
Nov 27 at 0:14
Small suggestion: the thing you're simplifying is an ellipse; an eclipse is an occultation of one celestial body by another.
– John Hughes
Nov 27 at 0:14
thank you.......
– Evan Kim
Nov 27 at 0:15
thank you.......
– Evan Kim
Nov 27 at 0:15
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
By writing
$$
begin{bmatrix}x\yend{bmatrix} = begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix} begin{bmatrix}x'\y'end{bmatrix}$$
(or the reverse), and mutliplying out, you get something that looks like
$$
x = ax' + b y'\
y = c x' + d y'
$$
where $a,b,c,d$ are things involving $sqrt{5}$ and other stuff.
You are now supposed to substitute those into the terms of
$$
2x^2 - 4xy + 5y^2 = 1
$$
to get something like
$$
2(ax' + by')^2 - 4ldots = 1
$$
which you expand out to get a quadratic expression in $x'$ and $y'$ rather than in $x$ and $y$.
I'm not going to do the work for you, but at least you now know what you're being asked to do. The idea was to do this stuff for a general $x',y'$ pair, not a particular numerical pair.
answered Nov 27 at 0:18
John Hughes
62.1k24090
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
By writing
$$
begin{bmatrix}x\yend{bmatrix} = begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix} begin{bmatrix}x'\y'end{bmatrix}$$
(or the reverse), and mutliplying out, you get something that looks like
$$
x = ax' + b y'\
y = c x' + d y'
$$
where $a,b,c,d$ are things involving $sqrt{5}$ and other stuff.
You are now supposed to substitute those into the terms of
$$
2x^2 - 4xy + 5y^2 = 1
$$
to get something like
$$
2(ax' + by')^2 - 4ldots = 1
$$
which you expand out to get a quadratic expression in $x'$ and $y'$ rather than in $x$ and $y$.
I'm not going to do the work for you, but at least you now know what you're being asked to do. The idea was to do this stuff for a general $x',y'$ pair, not a particular numerical pair.
answered Nov 27 at 0:18
John Hughes
62.1k24090
add a comment |
up vote
2
down vote
accepted
By writing
$$
begin{bmatrix}x\yend{bmatrix} = begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix} begin{bmatrix}x'\y'end{bmatrix}$$
(or the reverse), and mutliplying out, you get something that looks like
$$
x = ax' + b y'\
y = c x' + d y'
$$
where $a,b,c,d$ are things involving $sqrt{5}$ and other stuff.
You are now supposed to substitute those into the terms of
$$
2x^2 - 4xy + 5y^2 = 1
$$
to get something like
$$
2(ax' + by')^2 - 4ldots = 1
$$
which you expand out to get a quadratic expression in $x'$ and $y'$ rather than in $x$ and $y$.
I'm not going to do the work for you, but at least you now know what you're being asked to do. The idea was to do this stuff for a general $x',y'$ pair, not a particular numerical pair.
answered Nov 27 at 0:18
John Hughes
62.1k24090
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
By writing
$$
begin{bmatrix}x\yend{bmatrix} = begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix} begin{bmatrix}x'\y'end{bmatrix}$$
(or the reverse), and mutliplying out, you get something that looks like
$$
x = ax' + b y'\
y = c x' + d y'
$$
where $a,b,c,d$ are things involving $sqrt{5}$ and other stuff.
You are now supposed to substitute those into the terms of
$$
2x^2 - 4xy + 5y^2 = 1
$$
to get something like
$$
2(ax' + by')^2 - 4ldots = 1
$$
which you expand out to get a quadratic expression in $x'$ and $y'$ rather than in $x$ and $y$.
I'm not going to do the work for you, but at least you now know what you're being asked to do. The idea was to do this stuff for a general $x',y'$ pair, not a particular numerical pair.
answered Nov 27 at 0:18
John Hughes
62.1k24090
By writing
$$
begin{bmatrix}x\yend{bmatrix} = begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix} begin{bmatrix}x'\y'end{bmatrix}$$
(or the reverse), and mutliplying out, you get something that looks like
$$
x = ax' + b y'\
y = c x' + d y'
$$
where $a,b,c,d$ are things involving $sqrt{5}$ and other stuff.
You are now supposed to substitute those into the terms of
$$
2x^2 - 4xy + 5y^2 = 1
$$
to get something like
$$
2(ax' + by')^2 - 4ldots = 1
$$
which you expand out to get a quadratic expression in $x'$ and $y'$ rather than in $x$ and $y$.
I'm not going to do the work for you, but at least you now know what you're being asked to do. The idea was to do this stuff for a general $x',y'$ pair, not a particular numerical pair.
answered Nov 27 at 0:18
John Hughes
62.1k24090
answered Nov 27 at 0:18
John Hughes
62.1k24090
answered Nov 27 at 0:18
John Hughes
62.1k24090
answered Nov 27 at 0:18
John Hughes
62.1k24090
62.1k24090
add a comment |
add a comment |
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Small suggestion: the thing you're simplifying is an ellipse; an eclipse is an occultation of one celestial body by another.
– John Hughes
Nov 27 at 0:14
thank you.......
– Evan Kim
Nov 27 at 0:15