How do I change this parametric equation: $x=t+1/t, y=t^2 + 1/t^2$ into a Cartesian equation?
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I've just started parametric equations on my own & I am a bit confused on how to convert this parametric equation into a Cartesian equation.
$$beginarrayrcl
x=t + frac1t,
y= t^2 + frac1t^2
endarrayqquad$$
parametric curves
edited Aug 24 at 8:02
Rodrigo de Azevedo
12.7k41751
asked Apr 15 '12 at 11:49
Jallah
7319
add a comment |Â
up vote
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I've just started parametric equations on my own & I am a bit confused on how to convert this parametric equation into a Cartesian equation.
$$beginarrayrcl
x=t + frac1t,
y= t^2 + frac1t^2
endarrayqquad$$
parametric curves
edited Aug 24 at 8:02
Rodrigo de Azevedo
12.7k41751
asked Apr 15 '12 at 11:49
Jallah
7319
3
Add and subtract 2, $y = left(t+frac1tright)^2 - 2$.
– Ishaan Singh
Apr 15 '12 at 11:51
@IshaanSingh Please write those comments that answer a question as answer unless you think that the OP might have had something non-trivial to ask but turned out to be trivial because of a typo or other reasons why you think a comment is better than an answer. Here, I don't see any such--please correct me if I am wrong. Regards,
– user21436
Apr 15 '12 at 13:00
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've just started parametric equations on my own & I am a bit confused on how to convert this parametric equation into a Cartesian equation.
$$beginarrayrcl
x=t + frac1t,
y= t^2 + frac1t^2
endarrayqquad$$
parametric curves
edited Aug 24 at 8:02
Rodrigo de Azevedo
12.7k41751
asked Apr 15 '12 at 11:49
Jallah
7319
I've just started parametric equations on my own & I am a bit confused on how to convert this parametric equation into a Cartesian equation.
$$beginarrayrcl
x=t + frac1t,
y= t^2 + frac1t^2
endarrayqquad$$
parametric curves
edited Aug 24 at 8:02
Rodrigo de Azevedo
12.7k41751
asked Apr 15 '12 at 11:49
Jallah
7319
edited Aug 24 at 8:02
Rodrigo de Azevedo
12.7k41751
edited Aug 24 at 8:02
Rodrigo de Azevedo
12.7k41751
edited Aug 24 at 8:02
Rodrigo de Azevedo
12.7k41751
12.7k41751
asked Apr 15 '12 at 11:49
Jallah
7319
asked Apr 15 '12 at 11:49
Jallah
7319
asked Apr 15 '12 at 11:49
Jallah
7319
7319
3
Add and subtract 2, $y = left(t+frac1tright)^2 - 2$.
– Ishaan Singh
Apr 15 '12 at 11:51
@IshaanSingh Please write those comments that answer a question as answer unless you think that the OP might have had something non-trivial to ask but turned out to be trivial because of a typo or other reasons why you think a comment is better than an answer. Here, I don't see any such--please correct me if I am wrong. Regards,
– user21436
Apr 15 '12 at 13:00
add a comment |Â
3
Add and subtract 2, $y = left(t+frac1tright)^2 - 2$.
– Ishaan Singh
Apr 15 '12 at 11:51
@IshaanSingh Please write those comments that answer a question as answer unless you think that the OP might have had something non-trivial to ask but turned out to be trivial because of a typo or other reasons why you think a comment is better than an answer. Here, I don't see any such--please correct me if I am wrong. Regards,
– user21436
Apr 15 '12 at 13:00
3
3
Add and subtract 2, $y = left(t+frac1tright)^2 - 2$.
– Ishaan Singh
Apr 15 '12 at 11:51
Add and subtract 2, $y = left(t+frac1tright)^2 - 2$.
– Ishaan Singh
Apr 15 '12 at 11:51
@IshaanSingh Please write those comments that answer a question as answer unless you think that the OP might have had something non-trivial to ask but turned out to be trivial because of a typo or other reasons why you think a comment is better than an answer. Here, I don't see any such--please correct me if I am wrong. Regards,
– user21436
Apr 15 '12 at 13:00
@IshaanSingh Please write those comments that answer a question as answer unless you think that the OP might have had something non-trivial to ask but turned out to be trivial because of a typo or other reasons why you think a comment is better than an answer. Here, I don't see any such--please correct me if I am wrong. Regards,
– user21436
Apr 15 '12 at 13:00
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Hint: compute $x^2$ and subtract $y$
answered Apr 15 '12 at 11:51
Raymond Manzoni
37.2k562114
add a comment |Â
up vote
2
down vote
$$x=t+1/t$$
$$x^2=(t+1/t)^2$$
Expanding the perfect square $a^2+2ab+b^2$:
$$x^2=t^2+1/t^2+2$$
As $y=t^2+1/t^2$, therefore $y=x^2-2$
edited Jul 31 '17 at 2:40
user3658307
4,2393945
answered Jul 31 '17 at 2:01
Emma
211
Thanks for your answer. Please, however, use MathJax/latex next time :)
– user3658307
Jul 31 '17 at 2:42
add a comment |Â
up vote
-2
down vote
$$beginalign
x&=t+1/t\
xt-t&=1 \
t(x-1)&=1\
\
t&=1/(x-1)\
\
y&=t^2+1/t^2\
\
y&=1/(x-1)^2 + (x-1)^2
endalign$$
edited Jan 4 '14 at 19:05
K. Rmth
1,439722
answered Jan 4 '14 at 18:58
Helen
5
1
Your solution is not correct: the very first step is wrong. If $x = t + 1/t$, then multiplying both sides by $t$ gives $xt = colorredt^2 + 1$, not $xt = t+1$.
– heropup
Jan 4 '14 at 19:04
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: compute $x^2$ and subtract $y$
answered Apr 15 '12 at 11:51
Raymond Manzoni
37.2k562114
add a comment |Â
up vote
2
down vote
accepted
Hint: compute $x^2$ and subtract $y$
answered Apr 15 '12 at 11:51
Raymond Manzoni
37.2k562114
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: compute $x^2$ and subtract $y$
answered Apr 15 '12 at 11:51
Raymond Manzoni
37.2k562114
Hint: compute $x^2$ and subtract $y$
answered Apr 15 '12 at 11:51
Raymond Manzoni
37.2k562114
answered Apr 15 '12 at 11:51
Raymond Manzoni
37.2k562114
answered Apr 15 '12 at 11:51
Raymond Manzoni
37.2k562114
answered Apr 15 '12 at 11:51
Raymond Manzoni
37.2k562114
37.2k562114
add a comment |Â
add a comment |Â
up vote
2
down vote
$$x=t+1/t$$
$$x^2=(t+1/t)^2$$
Expanding the perfect square $a^2+2ab+b^2$:
$$x^2=t^2+1/t^2+2$$
As $y=t^2+1/t^2$, therefore $y=x^2-2$
edited Jul 31 '17 at 2:40
user3658307
4,2393945
answered Jul 31 '17 at 2:01
Emma
211
Thanks for your answer. Please, however, use MathJax/latex next time :)
– user3658307
Jul 31 '17 at 2:42
add a comment |Â
up vote
2
down vote
$$x=t+1/t$$
$$x^2=(t+1/t)^2$$
Expanding the perfect square $a^2+2ab+b^2$:
$$x^2=t^2+1/t^2+2$$
As $y=t^2+1/t^2$, therefore $y=x^2-2$
edited Jul 31 '17 at 2:40
user3658307
4,2393945
answered Jul 31 '17 at 2:01
Emma
211
Thanks for your answer. Please, however, use MathJax/latex next time :)
– user3658307
Jul 31 '17 at 2:42
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$x=t+1/t$$
$$x^2=(t+1/t)^2$$
Expanding the perfect square $a^2+2ab+b^2$:
$$x^2=t^2+1/t^2+2$$
As $y=t^2+1/t^2$, therefore $y=x^2-2$
edited Jul 31 '17 at 2:40
user3658307
4,2393945
answered Jul 31 '17 at 2:01
Emma
211
$$x=t+1/t$$
$$x^2=(t+1/t)^2$$
Expanding the perfect square $a^2+2ab+b^2$:
$$x^2=t^2+1/t^2+2$$
As $y=t^2+1/t^2$, therefore $y=x^2-2$
edited Jul 31 '17 at 2:40
user3658307
4,2393945
answered Jul 31 '17 at 2:01
Emma
211
edited Jul 31 '17 at 2:40
user3658307
4,2393945
edited Jul 31 '17 at 2:40
user3658307
4,2393945
edited Jul 31 '17 at 2:40
user3658307
4,2393945
4,2393945
answered Jul 31 '17 at 2:01
Emma
211
answered Jul 31 '17 at 2:01
Emma
211
answered Jul 31 '17 at 2:01
Emma
211
211
Thanks for your answer. Please, however, use MathJax/latex next time :)
– user3658307
Jul 31 '17 at 2:42
add a comment |Â
Thanks for your answer. Please, however, use MathJax/latex next time :)
– user3658307
Jul 31 '17 at 2:42
Thanks for your answer. Please, however, use MathJax/latex next time :)
– user3658307
Jul 31 '17 at 2:42
Thanks for your answer. Please, however, use MathJax/latex next time :)
– user3658307
Jul 31 '17 at 2:42
add a comment |Â
up vote
-2
down vote
$$beginalign
x&=t+1/t\
xt-t&=1 \
t(x-1)&=1\
\
t&=1/(x-1)\
\
y&=t^2+1/t^2\
\
y&=1/(x-1)^2 + (x-1)^2
endalign$$
edited Jan 4 '14 at 19:05
K. Rmth
1,439722
answered Jan 4 '14 at 18:58
Helen
5
1
Your solution is not correct: the very first step is wrong. If $x = t + 1/t$, then multiplying both sides by $t$ gives $xt = colorredt^2 + 1$, not $xt = t+1$.
– heropup
Jan 4 '14 at 19:04
add a comment |Â
up vote
-2
down vote
$$beginalign
x&=t+1/t\
xt-t&=1 \
t(x-1)&=1\
\
t&=1/(x-1)\
\
y&=t^2+1/t^2\
\
y&=1/(x-1)^2 + (x-1)^2
endalign$$
edited Jan 4 '14 at 19:05
K. Rmth
1,439722
answered Jan 4 '14 at 18:58
Helen
5
1
Your solution is not correct: the very first step is wrong. If $x = t + 1/t$, then multiplying both sides by $t$ gives $xt = colorredt^2 + 1$, not $xt = t+1$.
– heropup
Jan 4 '14 at 19:04
add a comment |Â
up vote
-2
down vote
up vote
-2
down vote
$$beginalign
x&=t+1/t\
xt-t&=1 \
t(x-1)&=1\
\
t&=1/(x-1)\
\
y&=t^2+1/t^2\
\
y&=1/(x-1)^2 + (x-1)^2
endalign$$
edited Jan 4 '14 at 19:05
K. Rmth
1,439722
answered Jan 4 '14 at 18:58
Helen
5
$$beginalign
x&=t+1/t\
xt-t&=1 \
t(x-1)&=1\
\
t&=1/(x-1)\
\
y&=t^2+1/t^2\
\
y&=1/(x-1)^2 + (x-1)^2
endalign$$
edited Jan 4 '14 at 19:05
K. Rmth
1,439722
answered Jan 4 '14 at 18:58
Helen
5
edited Jan 4 '14 at 19:05
K. Rmth
1,439722
edited Jan 4 '14 at 19:05
K. Rmth
1,439722
edited Jan 4 '14 at 19:05
K. Rmth
1,439722
1,439722
answered Jan 4 '14 at 18:58
Helen
5
answered Jan 4 '14 at 18:58
Helen
5
answered Jan 4 '14 at 18:58
Helen
5
5
1
Your solution is not correct: the very first step is wrong. If $x = t + 1/t$, then multiplying both sides by $t$ gives $xt = colorredt^2 + 1$, not $xt = t+1$.
– heropup
Jan 4 '14 at 19:04
add a comment |Â
1
Your solution is not correct: the very first step is wrong. If $x = t + 1/t$, then multiplying both sides by $t$ gives $xt = colorredt^2 + 1$, not $xt = t+1$.
– heropup
Jan 4 '14 at 19:04
1
1
Your solution is not correct: the very first step is wrong. If $x = t + 1/t$, then multiplying both sides by $t$ gives $xt = colorredt^2 + 1$, not $xt = t+1$.
– heropup
Jan 4 '14 at 19:04
Your solution is not correct: the very first step is wrong. If $x = t + 1/t$, then multiplying both sides by $t$ gives $xt = colorredt^2 + 1$, not $xt = t+1$.
– heropup
Jan 4 '14 at 19:04
add a comment |Â
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3
Add and subtract 2, $y = left(t+frac1tright)^2 - 2$.
– Ishaan Singh
Apr 15 '12 at 11:51
@IshaanSingh Please write those comments that answer a question as answer unless you think that the OP might have had something non-trivial to ask but turned out to be trivial because of a typo or other reasons why you think a comment is better than an answer. Here, I don't see any such--please correct me if I am wrong. Regards,
– user21436
Apr 15 '12 at 13:00