How to convert parametric equations to Cartesian form?

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I've been working on converting parametric equations into Cartesian form, but can't figure this out.



$$x = fract^2+1t^2-1$$
$$y = frac2tt^2-1$$



How do I covert that to Cartesian? Any help would be most appreciated.




parametric curves




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    up vote
    1
    down vote

    favorite












    I've been working on converting parametric equations into Cartesian form, but can't figure this out.



    $$x = fract^2+1t^2-1$$
    $$y = frac2tt^2-1$$



    How do I covert that to Cartesian? Any help would be most appreciated.




    parametric curves




    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I've been working on converting parametric equations into Cartesian form, but can't figure this out.



      $$x = fract^2+1t^2-1$$
      $$y = frac2tt^2-1$$



      How do I covert that to Cartesian? Any help would be most appreciated.




      parametric curves




      share|cite|improve this question














      I've been working on converting parametric equations into Cartesian form, but can't figure this out.



      $$x = fract^2+1t^2-1$$
      $$y = frac2tt^2-1$$



      How do I covert that to Cartesian? Any help would be most appreciated.





      parametric curves





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 24 at 7:59









      Rodrigo de Azevedo

      12.7k41751




      12.7k41751










      asked Feb 5 '15 at 22:13









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          2 Answers
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          Since $(t^2 - 1)x = t^2 + 1$ and $(t^2 - 1)y = 2t$, $$(t^2 - 1)^2 x^2 - (t^2 - 1)^2 y^2 = (t^2 + 1)^2 - 4t^2 = (t^4 + 2t^2 + 1) - 4t^2 = t^4 - 2t^2 + 1 = (t^2 - 1)^2.$$



          Dividing through by $(t^2 - 1)^2$, we obtain $$x^2 - y^2 = 1$$






          share|cite|improve this answer



























            up vote
            0
            down vote













            In general, you need to eliminate $t$ with the 2 equations.



            For this one, we get a shortcut.



            $$x+y=t+1 over t-1$$
            $$x-y=t-1 over t+1$$



            Then, obviously,



            $$(x+y)(x-y)=1$$
            $$(x^2-y^2)=1$$






            share|cite|improve this answer




















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              2 Answers
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              2 Answers
              2






              active

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              active

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              up vote
              1
              down vote













              Since $(t^2 - 1)x = t^2 + 1$ and $(t^2 - 1)y = 2t$, $$(t^2 - 1)^2 x^2 - (t^2 - 1)^2 y^2 = (t^2 + 1)^2 - 4t^2 = (t^4 + 2t^2 + 1) - 4t^2 = t^4 - 2t^2 + 1 = (t^2 - 1)^2.$$



              Dividing through by $(t^2 - 1)^2$, we obtain $$x^2 - y^2 = 1$$






              share|cite|improve this answer
























                up vote
                1
                down vote













                Since $(t^2 - 1)x = t^2 + 1$ and $(t^2 - 1)y = 2t$, $$(t^2 - 1)^2 x^2 - (t^2 - 1)^2 y^2 = (t^2 + 1)^2 - 4t^2 = (t^4 + 2t^2 + 1) - 4t^2 = t^4 - 2t^2 + 1 = (t^2 - 1)^2.$$



                Dividing through by $(t^2 - 1)^2$, we obtain $$x^2 - y^2 = 1$$






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Since $(t^2 - 1)x = t^2 + 1$ and $(t^2 - 1)y = 2t$, $$(t^2 - 1)^2 x^2 - (t^2 - 1)^2 y^2 = (t^2 + 1)^2 - 4t^2 = (t^4 + 2t^2 + 1) - 4t^2 = t^4 - 2t^2 + 1 = (t^2 - 1)^2.$$



                  Dividing through by $(t^2 - 1)^2$, we obtain $$x^2 - y^2 = 1$$






                  share|cite|improve this answer












                  Since $(t^2 - 1)x = t^2 + 1$ and $(t^2 - 1)y = 2t$, $$(t^2 - 1)^2 x^2 - (t^2 - 1)^2 y^2 = (t^2 + 1)^2 - 4t^2 = (t^4 + 2t^2 + 1) - 4t^2 = t^4 - 2t^2 + 1 = (t^2 - 1)^2.$$



                  Dividing through by $(t^2 - 1)^2$, we obtain $$x^2 - y^2 = 1$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 5 '15 at 22:34









                  kobe

                  34k22146




                  34k22146




















                      up vote
                      0
                      down vote













                      In general, you need to eliminate $t$ with the 2 equations.



                      For this one, we get a shortcut.



                      $$x+y=t+1 over t-1$$
                      $$x-y=t-1 over t+1$$



                      Then, obviously,



                      $$(x+y)(x-y)=1$$
                      $$(x^2-y^2)=1$$






                      share|cite|improve this answer
























                        up vote
                        0
                        down vote













                        In general, you need to eliminate $t$ with the 2 equations.



                        For this one, we get a shortcut.



                        $$x+y=t+1 over t-1$$
                        $$x-y=t-1 over t+1$$



                        Then, obviously,



                        $$(x+y)(x-y)=1$$
                        $$(x^2-y^2)=1$$






                        share|cite|improve this answer






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          In general, you need to eliminate $t$ with the 2 equations.



                          For this one, we get a shortcut.



                          $$x+y=t+1 over t-1$$
                          $$x-y=t-1 over t+1$$



                          Then, obviously,



                          $$(x+y)(x-y)=1$$
                          $$(x^2-y^2)=1$$






                          share|cite|improve this answer












                          In general, you need to eliminate $t$ with the 2 equations.



                          For this one, we get a shortcut.



                          $$x+y=t+1 over t-1$$
                          $$x-y=t-1 over t+1$$



                          Then, obviously,



                          $$(x+y)(x-y)=1$$
                          $$(x^2-y^2)=1$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Feb 5 '15 at 22:40









                          PdotWang

                          684411




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