Proving a point lies on a ellipse












1















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





























    1















    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



























      1












      1








      1


      0






      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 '18 at 11:48









      Ng Chung Tak

      14.2k31334




      14.2k31334










      asked Nov 22 '18 at 11:15









      Matlab rookieMatlab rookie

      357




      357






















          1 Answer
          1






          active

          oldest

          votes


















          1














          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























            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',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            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%2f3009014%2fproving-a-point-lies-on-a-ellipse%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            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




























              1














              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


























                1












                1








                1






                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 '18 at 11:41

























                answered Nov 22 '18 at 11:34









                Jean-Claude ArbautJean-Claude Arbaut

                14.7k63464




                14.7k63464






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009014%2fproving-a-point-lies-on-a-ellipse%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                    Image
                    28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                    Read more

                    ts Property 'filter' does not exist on type '{}'

                    Image
                    1 I'm trying to follow Angular Material guide to create a custom filter for autocomplete input text https://material.angular.io/components/autocomplete/overview This is what I have TS import { Component, OnInit } from '@angular/core'; import { TemplateRef } from '@angular/core'; import { FormControl } from "@angular/forms"; import { Observable } from "rxjs"; import { map, startWith, filter } from 'rxjs/operators'; (...) export class AppComponent implements OnInit { title = 'Your Profile'; displayedColumns: string = ['selected', 'name', 'description', 'pnr']; dataSource = ELEMENT_DATA; myControl = new FormControl(); options: any = ['One', 'Two', 'Three']; filteredOptions:
                    Read more

                    mat-slide-toggle shouldn't change it's state when I click cancel in confirmation window

                    Image
                    1 When I checked the slide toggle it should work as expected. But when I try to uncheck it there is a confirmation window asking if you are sure. When I click cancel in that confirmation window, still the toggle changes to unchecked, which shouldn't happen. It should stay in the same state. Here is my Html and TS Code // html <mat-slide-toggle (change)="change($event)" [checked]="isChecked()" > To-pay </mat-slide-toggle> TS code: // ts change(e) { if(this.checked) { if(confirm("Are you sure")) { console.log("toggle") } else { e.stopPropagation(); console.log("toggle should not change if I click the cancel button") }
                    Read more