How does one explicitly show that the distance from Focus_1 to the ellipse plus the distance from Focus_2 to...












0














When deriving the equation for an ellipse, I have seen that one of the first steps is to recognize that for a point on the ellipse defined by y=0, the two distances from Focus_1 (call it D1) and Focus_2 (call it D2) to the ellipse add up to 2*a, where 'a' is the length from the origin to the ellipse at y = 0.



Logically, as long as you preface this by stating that "If there are two foci that are equidistant from the origin..." then this makes sense.



However, I am then confused when this equivalency (D1 + D2 = 2*a) is subsequently used for the remainder of the derivation. D1 changes its value as you move around the ellipse (i.e. when you're no longer at y = 0); D2, of course, changes its value as well.



Since D1 + D2 = 2a was only determined for the condition that y=0, why can I subsequently use this relationship for different y values on this ellipse. For all I know, D1 + D2 does not equal 2*a for any other y value. It seems like a logical inconsistency.



Am I thinking about this incorrectly?






geometry logic conic-sections







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    When deriving the equation for an ellipse, I have seen that one of the first steps is to recognize that for a point on the ellipse defined by y=0, the two distances from Focus_1 (call it D1) and Focus_2 (call it D2) to the ellipse add up to 2*a, where 'a' is the length from the origin to the ellipse at y = 0.



    Logically, as long as you preface this by stating that "If there are two foci that are equidistant from the origin..." then this makes sense.



    However, I am then confused when this equivalency (D1 + D2 = 2*a) is subsequently used for the remainder of the derivation. D1 changes its value as you move around the ellipse (i.e. when you're no longer at y = 0); D2, of course, changes its value as well.



    Since D1 + D2 = 2a was only determined for the condition that y=0, why can I subsequently use this relationship for different y values on this ellipse. For all I know, D1 + D2 does not equal 2*a for any other y value. It seems like a logical inconsistency.



    Am I thinking about this incorrectly?






    geometry logic conic-sections







    share|cite|improve this question

























      0












      0








      0







      When deriving the equation for an ellipse, I have seen that one of the first steps is to recognize that for a point on the ellipse defined by y=0, the two distances from Focus_1 (call it D1) and Focus_2 (call it D2) to the ellipse add up to 2*a, where 'a' is the length from the origin to the ellipse at y = 0.



      Logically, as long as you preface this by stating that "If there are two foci that are equidistant from the origin..." then this makes sense.



      However, I am then confused when this equivalency (D1 + D2 = 2*a) is subsequently used for the remainder of the derivation. D1 changes its value as you move around the ellipse (i.e. when you're no longer at y = 0); D2, of course, changes its value as well.



      Since D1 + D2 = 2a was only determined for the condition that y=0, why can I subsequently use this relationship for different y values on this ellipse. For all I know, D1 + D2 does not equal 2*a for any other y value. It seems like a logical inconsistency.



      Am I thinking about this incorrectly?






      geometry logic conic-sections







      share|cite|improve this question













      When deriving the equation for an ellipse, I have seen that one of the first steps is to recognize that for a point on the ellipse defined by y=0, the two distances from Focus_1 (call it D1) and Focus_2 (call it D2) to the ellipse add up to 2*a, where 'a' is the length from the origin to the ellipse at y = 0.



      Logically, as long as you preface this by stating that "If there are two foci that are equidistant from the origin..." then this makes sense.



      However, I am then confused when this equivalency (D1 + D2 = 2*a) is subsequently used for the remainder of the derivation. D1 changes its value as you move around the ellipse (i.e. when you're no longer at y = 0); D2, of course, changes its value as well.



      Since D1 + D2 = 2a was only determined for the condition that y=0, why can I subsequently use this relationship for different y values on this ellipse. For all I know, D1 + D2 does not equal 2*a for any other y value. It seems like a logical inconsistency.



      Am I thinking about this incorrectly?






      geometry logic conic-sections




      geometry logic conic-sections






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      asked Jan 4 at 5:06









      S.CramerS.Cramer

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          The definition of the ellipse is the set of points whose total distance from the two fixed points called foci is a constant.



          Since this constant at one instance is $2a$ it has to be $2a$ for every other point on the ellipse as well.






          share|cite|improve this answer





















          • So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
            – S.Cramer
            Jan 4 at 5:14








          • 1




            yes, it is much easier to put the origin at the center of the ellipse.
            – Mohammad Riazi-Kermani
            Jan 4 at 5:16



















          0














          There extists some set of points $(x,y)$ shuch that the distance from this point to each focus is constant.



          $d((c,0),(x,y)) + d((-c,0),(x,y)) = 2a\
          sqrt{(x-c)^2 + y^2} + sqrt{(x+c)^2 + y^2} = 2a$



          Now we have some algebra to cruch through...Square both sides



          $(x-c)^2 + y^2 + (x+c)^2 + y^2 + 2sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 4a^2$



          Isolate the radical on one side and the terms not under the radical on the other.



          $sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 2a^2-x^2-y^2 -c^2$



          square both sides again.



          $(x^2 + y^2 +c^2- 2xc)(x^2 + y^2+ c^2 +2xc) = (2a^2 - x^2-y^2 - c^2)^2$



          Multiply the two sides out.
          $x^4 + y^4 +c^4 - 4x^2c^2 +2x^2y^2 +2x^2c^2 + 2y^2c^2= 4a^4+x^4+y^4+c^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2+ 2x^2y^2 + 2x^2 c^2 + 2y^2c^2$



          Subtract like terms from both sides.



          $- 4x^2c^2 = 4a^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2$



          Move the x and y terms to one side and the contant terms to the other.



          $4(a^2-c^2)x^2 + 4a^2y^2 = 4a^2(a^2-c^2)\
          frac {x^2}{a^2} + frac {y^2}{a^2-c^2} = 1$



          let $a^2-c^2 = b^2$



          $frac {x^2}{a^2} + frac {y^2}{b^2} = 1$






          share|cite|improve this answer





















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            The definition of the ellipse is the set of points whose total distance from the two fixed points called foci is a constant.



            Since this constant at one instance is $2a$ it has to be $2a$ for every other point on the ellipse as well.






            share|cite|improve this answer





















            • So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
              – S.Cramer
              Jan 4 at 5:14








            • 1




              yes, it is much easier to put the origin at the center of the ellipse.
              – Mohammad Riazi-Kermani
              Jan 4 at 5:16
















            1














            The definition of the ellipse is the set of points whose total distance from the two fixed points called foci is a constant.



            Since this constant at one instance is $2a$ it has to be $2a$ for every other point on the ellipse as well.






            share|cite|improve this answer





















            • So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
              – S.Cramer
              Jan 4 at 5:14








            • 1




              yes, it is much easier to put the origin at the center of the ellipse.
              – Mohammad Riazi-Kermani
              Jan 4 at 5:16














            1












            1








            1






            The definition of the ellipse is the set of points whose total distance from the two fixed points called foci is a constant.



            Since this constant at one instance is $2a$ it has to be $2a$ for every other point on the ellipse as well.






            share|cite|improve this answer












            The definition of the ellipse is the set of points whose total distance from the two fixed points called foci is a constant.



            Since this constant at one instance is $2a$ it has to be $2a$ for every other point on the ellipse as well.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 4 at 5:12









            Mohammad Riazi-KermaniMohammad Riazi-Kermani

            41k42059




            41k42059












            • So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
              – S.Cramer
              Jan 4 at 5:14








            • 1




              yes, it is much easier to put the origin at the center of the ellipse.
              – Mohammad Riazi-Kermani
              Jan 4 at 5:16


















            • So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
              – S.Cramer
              Jan 4 at 5:14








            • 1




              yes, it is much easier to put the origin at the center of the ellipse.
              – Mohammad Riazi-Kermani
              Jan 4 at 5:16
















            So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
            – S.Cramer
            Jan 4 at 5:14






            So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
            – S.Cramer
            Jan 4 at 5:14






            1




            1




            yes, it is much easier to put the origin at the center of the ellipse.
            – Mohammad Riazi-Kermani
            Jan 4 at 5:16




            yes, it is much easier to put the origin at the center of the ellipse.
            – Mohammad Riazi-Kermani
            Jan 4 at 5:16











            0














            There extists some set of points $(x,y)$ shuch that the distance from this point to each focus is constant.



            $d((c,0),(x,y)) + d((-c,0),(x,y)) = 2a\
            sqrt{(x-c)^2 + y^2} + sqrt{(x+c)^2 + y^2} = 2a$



            Now we have some algebra to cruch through...Square both sides



            $(x-c)^2 + y^2 + (x+c)^2 + y^2 + 2sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 4a^2$



            Isolate the radical on one side and the terms not under the radical on the other.



            $sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 2a^2-x^2-y^2 -c^2$



            square both sides again.



            $(x^2 + y^2 +c^2- 2xc)(x^2 + y^2+ c^2 +2xc) = (2a^2 - x^2-y^2 - c^2)^2$



            Multiply the two sides out.
            $x^4 + y^4 +c^4 - 4x^2c^2 +2x^2y^2 +2x^2c^2 + 2y^2c^2= 4a^4+x^4+y^4+c^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2+ 2x^2y^2 + 2x^2 c^2 + 2y^2c^2$



            Subtract like terms from both sides.



            $- 4x^2c^2 = 4a^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2$



            Move the x and y terms to one side and the contant terms to the other.



            $4(a^2-c^2)x^2 + 4a^2y^2 = 4a^2(a^2-c^2)\
            frac {x^2}{a^2} + frac {y^2}{a^2-c^2} = 1$



            let $a^2-c^2 = b^2$



            $frac {x^2}{a^2} + frac {y^2}{b^2} = 1$






            share|cite|improve this answer


























              0














              There extists some set of points $(x,y)$ shuch that the distance from this point to each focus is constant.



              $d((c,0),(x,y)) + d((-c,0),(x,y)) = 2a\
              sqrt{(x-c)^2 + y^2} + sqrt{(x+c)^2 + y^2} = 2a$



              Now we have some algebra to cruch through...Square both sides



              $(x-c)^2 + y^2 + (x+c)^2 + y^2 + 2sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 4a^2$



              Isolate the radical on one side and the terms not under the radical on the other.



              $sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 2a^2-x^2-y^2 -c^2$



              square both sides again.



              $(x^2 + y^2 +c^2- 2xc)(x^2 + y^2+ c^2 +2xc) = (2a^2 - x^2-y^2 - c^2)^2$



              Multiply the two sides out.
              $x^4 + y^4 +c^4 - 4x^2c^2 +2x^2y^2 +2x^2c^2 + 2y^2c^2= 4a^4+x^4+y^4+c^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2+ 2x^2y^2 + 2x^2 c^2 + 2y^2c^2$



              Subtract like terms from both sides.



              $- 4x^2c^2 = 4a^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2$



              Move the x and y terms to one side and the contant terms to the other.



              $4(a^2-c^2)x^2 + 4a^2y^2 = 4a^2(a^2-c^2)\
              frac {x^2}{a^2} + frac {y^2}{a^2-c^2} = 1$



              let $a^2-c^2 = b^2$



              $frac {x^2}{a^2} + frac {y^2}{b^2} = 1$






              share|cite|improve this answer
























                0












                0








                0






                There extists some set of points $(x,y)$ shuch that the distance from this point to each focus is constant.



                $d((c,0),(x,y)) + d((-c,0),(x,y)) = 2a\
                sqrt{(x-c)^2 + y^2} + sqrt{(x+c)^2 + y^2} = 2a$



                Now we have some algebra to cruch through...Square both sides



                $(x-c)^2 + y^2 + (x+c)^2 + y^2 + 2sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 4a^2$



                Isolate the radical on one side and the terms not under the radical on the other.



                $sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 2a^2-x^2-y^2 -c^2$



                square both sides again.



                $(x^2 + y^2 +c^2- 2xc)(x^2 + y^2+ c^2 +2xc) = (2a^2 - x^2-y^2 - c^2)^2$



                Multiply the two sides out.
                $x^4 + y^4 +c^4 - 4x^2c^2 +2x^2y^2 +2x^2c^2 + 2y^2c^2= 4a^4+x^4+y^4+c^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2+ 2x^2y^2 + 2x^2 c^2 + 2y^2c^2$



                Subtract like terms from both sides.



                $- 4x^2c^2 = 4a^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2$



                Move the x and y terms to one side and the contant terms to the other.



                $4(a^2-c^2)x^2 + 4a^2y^2 = 4a^2(a^2-c^2)\
                frac {x^2}{a^2} + frac {y^2}{a^2-c^2} = 1$



                let $a^2-c^2 = b^2$



                $frac {x^2}{a^2} + frac {y^2}{b^2} = 1$






                share|cite|improve this answer












                There extists some set of points $(x,y)$ shuch that the distance from this point to each focus is constant.



                $d((c,0),(x,y)) + d((-c,0),(x,y)) = 2a\
                sqrt{(x-c)^2 + y^2} + sqrt{(x+c)^2 + y^2} = 2a$



                Now we have some algebra to cruch through...Square both sides



                $(x-c)^2 + y^2 + (x+c)^2 + y^2 + 2sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 4a^2$



                Isolate the radical on one side and the terms not under the radical on the other.



                $sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 2a^2-x^2-y^2 -c^2$



                square both sides again.



                $(x^2 + y^2 +c^2- 2xc)(x^2 + y^2+ c^2 +2xc) = (2a^2 - x^2-y^2 - c^2)^2$



                Multiply the two sides out.
                $x^4 + y^4 +c^4 - 4x^2c^2 +2x^2y^2 +2x^2c^2 + 2y^2c^2= 4a^4+x^4+y^4+c^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2+ 2x^2y^2 + 2x^2 c^2 + 2y^2c^2$



                Subtract like terms from both sides.



                $- 4x^2c^2 = 4a^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2$



                Move the x and y terms to one side and the contant terms to the other.



                $4(a^2-c^2)x^2 + 4a^2y^2 = 4a^2(a^2-c^2)\
                frac {x^2}{a^2} + frac {y^2}{a^2-c^2} = 1$



                let $a^2-c^2 = b^2$



                $frac {x^2}{a^2} + frac {y^2}{b^2} = 1$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 5:31









                Doug MDoug M

                44.2k31854




                44.2k31854






























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                    Display a custom attribute below product name in the front-end Magento 1.9.3.8

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                    0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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