How do I change this parametric equation: $x=t+1/t, y=t^2 + 1/t^2$ into a Cartesian equation?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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




share|cite|improve this question


















  • 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














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




share|cite|improve this question


















  • 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












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




share|cite|improve this question














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





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 24 at 8:02









Rodrigo de Azevedo

12.7k41751




12.7k41751










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












  • 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










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










Hint: compute $x^2$ and subtract $y$






share|cite|improve this answer



























    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 $y=x^2-2$






    share|cite|improve this answer






















    • Thanks for your answer. Please, however, use MathJax/latex next time :)
      – user3658307
      Jul 31 '17 at 2:42

















    up vote
    -2
    down vote













    $$beginalign
    x&=t+1/t\
    xt-t&=1 \
    t(x-1)&=1\
    \
    t&=1/(x-1)\
    \
    y&=t^2+1/t^2\
    \
    y&=1/(x-1)^2 + (x-1)^2
    endalign$$






    share|cite|improve this answer


















    • 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 Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f132051%2fhow-do-i-change-this-parametric-equation-x-t1-t-y-t2-1-t2-into-a-cartes%23new-answer', 'question_page');

    );

    Post as a guest

















    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 $y$






    share|cite|improve this answer
























      up vote
      2
      down vote



      accepted










      Hint: compute $x^2$ and subtract $y$






      share|cite|improve this answer






















        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Hint: compute $x^2$ and subtract $y$






        share|cite|improve this answer












        Hint: compute $x^2$ and subtract $y$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 15 '12 at 11:51









        Raymond Manzoni

        37.2k562114




        37.2k562114




















            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 $y=x^2-2$






            share|cite|improve this answer






















            • Thanks for your answer. Please, however, use MathJax/latex next time :)
              – user3658307
              Jul 31 '17 at 2:42














            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 $y=x^2-2$






            share|cite|improve this answer






















            • Thanks for your answer. Please, however, use MathJax/latex next time :)
              – user3658307
              Jul 31 '17 at 2:42












            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 $y=x^2-2$






            share|cite|improve this answer














            $$x=t+1/t$$



            $$x^2=(t+1/t)^2$$



            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 $y=x^2-2$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 31 '17 at 2:40









            user3658307

            4,2393945




            4,2393945










            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
















            • 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










            up vote
            -2
            down vote













            $$beginalign
            x&=t+1/t\
            xt-t&=1 \
            t(x-1)&=1\
            \
            t&=1/(x-1)\
            \
            y&=t^2+1/t^2\
            \
            y&=1/(x-1)^2 + (x-1)^2
            endalign$$






            share|cite|improve this answer


















            • 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














            up vote
            -2
            down vote













            $$beginalign
            x&=t+1/t\
            xt-t&=1 \
            t(x-1)&=1\
            \
            t&=1/(x-1)\
            \
            y&=t^2+1/t^2\
            \
            y&=1/(x-1)^2 + (x-1)^2
            endalign$$






            share|cite|improve this answer


















            • 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












            up vote
            -2
            down vote










            up vote
            -2
            down vote









            $$beginalign
            x&=t+1/t\
            xt-t&=1 \
            t(x-1)&=1\
            \
            t&=1/(x-1)\
            \
            y&=t^2+1/t^2\
            \
            y&=1/(x-1)^2 + (x-1)^2
            endalign$$






            share|cite|improve this answer














            $$beginalign
            x&=t+1/t\
            xt-t&=1 \
            t(x-1)&=1\
            \
            t&=1/(x-1)\
            \
            y&=t^2+1/t^2\
            \
            y&=1/(x-1)^2 + (x-1)^2
            endalign$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 4 '14 at 19:05









            K. Rmth

            1,439722




            1,439722










            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












            • 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

















             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f132051%2fhow-do-i-change-this-parametric-equation-x-t1-t-y-t2-1-t2-into-a-cartes%23new-answer', 'question_page');

            );

            Post as a guest
































































            -

            這個網誌中的熱門文章

            How to combine Bézier curves to a surface?

            圖片
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
            閱讀完整內容

            Mutual Information Always Non-negative

            圖片
            Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
            閱讀完整內容

            Why am i infinitely getting the same tweet with the Twitter Search API?

            圖片
            up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
            閱讀完整內容