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A curve is defined by the parametric equations $x=2t+frac{1}{t^2},; y=2t-frac{1}{t^2}$. Find the Cartesian...
$begingroup$
A curve is defined by the parametric equations
$$x=2t+frac{1}{t^2}$$
$$y=2t-frac{1}{t^2}$$
Show that the curve has the Cartesian equation $(x-y)(x+y)^2=k$
So I understand I need to eliminate the parameter $t$, but I'm not seeing an easy way to do this as I cannot rearrange for $t$ and then substitute. Any help will be appreciated.
parametric curves
edited Jan 20 at 14:35
user376343
3,9234829
asked Jan 20 at 12:01
H.LinkhornH.Linkhorn
473113
$endgroup$
add a comment |
$begingroup$
A curve is defined by the parametric equations
$$x=2t+frac{1}{t^2}$$
$$y=2t-frac{1}{t^2}$$
Show that the curve has the Cartesian equation $(x-y)(x+y)^2=k$
So I understand I need to eliminate the parameter $t$, but I'm not seeing an easy way to do this as I cannot rearrange for $t$ and then substitute. Any help will be appreciated.
parametric curves
edited Jan 20 at 14:35
user376343
3,9234829
asked Jan 20 at 12:01
H.LinkhornH.Linkhorn
473113
$endgroup$
add a comment |
$begingroup$
A curve is defined by the parametric equations
$$x=2t+frac{1}{t^2}$$
$$y=2t-frac{1}{t^2}$$
Show that the curve has the Cartesian equation $(x-y)(x+y)^2=k$
So I understand I need to eliminate the parameter $t$, but I'm not seeing an easy way to do this as I cannot rearrange for $t$ and then substitute. Any help will be appreciated.
parametric curves
edited Jan 20 at 14:35
user376343
3,9234829
asked Jan 20 at 12:01
H.LinkhornH.Linkhorn
473113
$endgroup$
A curve is defined by the parametric equations
$$x=2t+frac{1}{t^2}$$
$$y=2t-frac{1}{t^2}$$
Show that the curve has the Cartesian equation $(x-y)(x+y)^2=k$
So I understand I need to eliminate the parameter $t$, but I'm not seeing an easy way to do this as I cannot rearrange for $t$ and then substitute. Any help will be appreciated.
parametric curves
parametric curves
edited Jan 20 at 14:35
user376343
3,9234829
asked Jan 20 at 12:01
H.LinkhornH.Linkhorn
473113
edited Jan 20 at 14:35
user376343
3,9234829
asked Jan 20 at 12:01
H.LinkhornH.Linkhorn
473113
edited Jan 20 at 14:35
user376343
3,9234829
edited Jan 20 at 14:35
user376343
3,9234829
edited Jan 20 at 14:35
user376343
3,9234829
3,9234829
asked Jan 20 at 12:01
H.LinkhornH.Linkhorn
473113
asked Jan 20 at 12:01
H.LinkhornH.Linkhorn
473113
asked Jan 20 at 12:01
H.LinkhornH.Linkhorn
473113
473113
add a comment |
add a comment |
2 Answers
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$begingroup$
$$x-y=frac{2}{t^2}$$
$$(x+y)^2=(4t)^2=16t^2$$
$$(x-y)(x+y)^2=frac{2}{t^2}(16t^2)=32$$
answered Jan 20 at 12:11
LarryLarry
2,41331129
$endgroup$
add a comment |
$begingroup$
Whenever is given a parametric system $$begin{cases}x=a(t)+b(t)\y=a(t)-b(t),;end{cases}$$ computing $x+y;$ and $;x-y;$ can help to eliminate the parameter.
In the present case $$begin{cases}x+y=4t\x-y={2over{t^2}}end{cases}$$ Since $(x-y)=k(x+y)^{-2},$ we obtain the constant as $(x-y)(x+y)^2.$
answered Jan 20 at 14:45
user376343user376343
3,9234829
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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$begingroup$
$$x-y=frac{2}{t^2}$$
$$(x+y)^2=(4t)^2=16t^2$$
$$(x-y)(x+y)^2=frac{2}{t^2}(16t^2)=32$$
answered Jan 20 at 12:11
LarryLarry
2,41331129
$endgroup$
add a comment |
$begingroup$
$$x-y=frac{2}{t^2}$$
$$(x+y)^2=(4t)^2=16t^2$$
$$(x-y)(x+y)^2=frac{2}{t^2}(16t^2)=32$$
answered Jan 20 at 12:11
LarryLarry
2,41331129
$endgroup$
add a comment |
$begingroup$
$$x-y=frac{2}{t^2}$$
$$(x+y)^2=(4t)^2=16t^2$$
$$(x-y)(x+y)^2=frac{2}{t^2}(16t^2)=32$$
answered Jan 20 at 12:11
LarryLarry
2,41331129
$endgroup$
$$x-y=frac{2}{t^2}$$
$$(x+y)^2=(4t)^2=16t^2$$
$$(x-y)(x+y)^2=frac{2}{t^2}(16t^2)=32$$
answered Jan 20 at 12:11
LarryLarry
2,41331129
answered Jan 20 at 12:11
LarryLarry
2,41331129
answered Jan 20 at 12:11
LarryLarry
2,41331129
answered Jan 20 at 12:11
LarryLarry
2,41331129
2,41331129
add a comment |
add a comment |
$begingroup$
Whenever is given a parametric system $$begin{cases}x=a(t)+b(t)\y=a(t)-b(t),;end{cases}$$ computing $x+y;$ and $;x-y;$ can help to eliminate the parameter.
In the present case $$begin{cases}x+y=4t\x-y={2over{t^2}}end{cases}$$ Since $(x-y)=k(x+y)^{-2},$ we obtain the constant as $(x-y)(x+y)^2.$
answered Jan 20 at 14:45
user376343user376343
3,9234829
$endgroup$
add a comment |
$begingroup$
Whenever is given a parametric system $$begin{cases}x=a(t)+b(t)\y=a(t)-b(t),;end{cases}$$ computing $x+y;$ and $;x-y;$ can help to eliminate the parameter.
In the present case $$begin{cases}x+y=4t\x-y={2over{t^2}}end{cases}$$ Since $(x-y)=k(x+y)^{-2},$ we obtain the constant as $(x-y)(x+y)^2.$
answered Jan 20 at 14:45
user376343user376343
3,9234829
$endgroup$
add a comment |
$begingroup$
Whenever is given a parametric system $$begin{cases}x=a(t)+b(t)\y=a(t)-b(t),;end{cases}$$ computing $x+y;$ and $;x-y;$ can help to eliminate the parameter.
In the present case $$begin{cases}x+y=4t\x-y={2over{t^2}}end{cases}$$ Since $(x-y)=k(x+y)^{-2},$ we obtain the constant as $(x-y)(x+y)^2.$
answered Jan 20 at 14:45
user376343user376343
3,9234829
$endgroup$
Whenever is given a parametric system $$begin{cases}x=a(t)+b(t)\y=a(t)-b(t),;end{cases}$$ computing $x+y;$ and $;x-y;$ can help to eliminate the parameter.
In the present case $$begin{cases}x+y=4t\x-y={2over{t^2}}end{cases}$$ Since $(x-y)=k(x+y)^{-2},$ we obtain the constant as $(x-y)(x+y)^2.$
answered Jan 20 at 14:45
user376343user376343
3,9234829
answered Jan 20 at 14:45
user376343user376343
3,9234829
answered Jan 20 at 14:45
user376343user376343
3,9234829
answered Jan 20 at 14:45
user376343user376343
3,9234829
3,9234829
add a comment |
add a comment |
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