Proving a point lies on a ellipse











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Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.




Hint given is to consider $2$ similar right angle triangles.



enter image description here



I presume that is reference to the $2$ triangles in the diagram but am not sure. All help appreciated.






geometry conic-sections







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    Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.




    Hint given is to consider $2$ similar right angle triangles.



    enter image description here



    I presume that is reference to the $2$ triangles in the diagram but am not sure. All help appreciated.






    geometry conic-sections







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.




      Hint given is to consider $2$ similar right angle triangles.



      enter image description here



      I presume that is reference to the $2$ triangles in the diagram but am not sure. All help appreciated.






      geometry conic-sections







      share|cite|improve this question
















      Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.




      Hint given is to consider $2$ similar right angle triangles.



      enter image description here



      I presume that is reference to the $2$ triangles in the diagram but am not sure. All help appreciated.






      geometry conic-sections




      geometry conic-sections






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      edited Nov 22 at 11:48









      Ng Chung Tak

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      13.8k31234










      asked Nov 22 at 11:15









      Matlab rookie

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          Let $O$, the center of the ellipse, be the origin of coordinates.
          Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
          Then $X^2+Y^2=CD^2=(a+b)^2$.



          Use Thales' theorem in triangles $PHC$ and $DOC$:



          $$frac{X-x}X=frac{b}{a+b}$$



          $$x=X-frac{bX}{a+b}=frac{aX}{a+b}$$



          Likewise in triangles $DKP$ and $DOC$:



          $$frac{Y-y}{Y}=frac{a}{a+b}$$
          $$y=Y-frac{aY}{a+b}=frac{bY}{a+b}$$



          Now,



          $$frac{x^2}{a^2}+frac{y^2}{b^2}=frac{X^2+Y^2}{(a+b)^2}=1$$



          And the point $P$ lies on the ellipse of equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.



          This property gives a very simple method to draw an ellipse: the paper strip method.






          share|cite|improve this answer























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            1 Answer
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            active

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            up vote
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            accepted










            Let $O$, the center of the ellipse, be the origin of coordinates.
            Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
            Then $X^2+Y^2=CD^2=(a+b)^2$.



            Use Thales' theorem in triangles $PHC$ and $DOC$:



            $$frac{X-x}X=frac{b}{a+b}$$



            $$x=X-frac{bX}{a+b}=frac{aX}{a+b}$$



            Likewise in triangles $DKP$ and $DOC$:



            $$frac{Y-y}{Y}=frac{a}{a+b}$$
            $$y=Y-frac{aY}{a+b}=frac{bY}{a+b}$$



            Now,



            $$frac{x^2}{a^2}+frac{y^2}{b^2}=frac{X^2+Y^2}{(a+b)^2}=1$$



            And the point $P$ lies on the ellipse of equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.



            This property gives a very simple method to draw an ellipse: the paper strip method.






            share|cite|improve this answer



























              up vote
              1
              down vote



              accepted










              Let $O$, the center of the ellipse, be the origin of coordinates.
              Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
              Then $X^2+Y^2=CD^2=(a+b)^2$.



              Use Thales' theorem in triangles $PHC$ and $DOC$:



              $$frac{X-x}X=frac{b}{a+b}$$



              $$x=X-frac{bX}{a+b}=frac{aX}{a+b}$$



              Likewise in triangles $DKP$ and $DOC$:



              $$frac{Y-y}{Y}=frac{a}{a+b}$$
              $$y=Y-frac{aY}{a+b}=frac{bY}{a+b}$$



              Now,



              $$frac{x^2}{a^2}+frac{y^2}{b^2}=frac{X^2+Y^2}{(a+b)^2}=1$$



              And the point $P$ lies on the ellipse of equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.



              This property gives a very simple method to draw an ellipse: the paper strip method.






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Let $O$, the center of the ellipse, be the origin of coordinates.
                Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
                Then $X^2+Y^2=CD^2=(a+b)^2$.



                Use Thales' theorem in triangles $PHC$ and $DOC$:



                $$frac{X-x}X=frac{b}{a+b}$$



                $$x=X-frac{bX}{a+b}=frac{aX}{a+b}$$



                Likewise in triangles $DKP$ and $DOC$:



                $$frac{Y-y}{Y}=frac{a}{a+b}$$
                $$y=Y-frac{aY}{a+b}=frac{bY}{a+b}$$



                Now,



                $$frac{x^2}{a^2}+frac{y^2}{b^2}=frac{X^2+Y^2}{(a+b)^2}=1$$



                And the point $P$ lies on the ellipse of equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.



                This property gives a very simple method to draw an ellipse: the paper strip method.






                share|cite|improve this answer














                Let $O$, the center of the ellipse, be the origin of coordinates.
                Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$.
                Then $X^2+Y^2=CD^2=(a+b)^2$.



                Use Thales' theorem in triangles $PHC$ and $DOC$:



                $$frac{X-x}X=frac{b}{a+b}$$



                $$x=X-frac{bX}{a+b}=frac{aX}{a+b}$$



                Likewise in triangles $DKP$ and $DOC$:



                $$frac{Y-y}{Y}=frac{a}{a+b}$$
                $$y=Y-frac{aY}{a+b}=frac{bY}{a+b}$$



                Now,



                $$frac{x^2}{a^2}+frac{y^2}{b^2}=frac{X^2+Y^2}{(a+b)^2}=1$$



                And the point $P$ lies on the ellipse of equation $frac{x^2}{a^2}+frac{y^2}{b^2}=1$.



                This property gives a very simple method to draw an ellipse: the paper strip method.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 22 at 11:41

























                answered Nov 22 at 11:34









                Jean-Claude Arbaut

                14.7k63363




                14.7k63363






























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