If $y=mx + c$ is a tangent to an ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$, then $c^2=a^2m^2 + b^2$












0















If $y=mx + c$ is a tangent to an ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$, then show that $c^2=a^2m^2 + b^2$.




So for this question, first off I tried to differentiate it using implicit differentiation. I got
$$frac{dy}{dx}=-frac{xb^2}{ya^2}$$



Afterward, I was unsure how to proceed. I tried subbing the above into $y=mx +c$, but I don't think that helps. How to I get $c^2=a^2m^2 + b^2$ out?






conic-sections







share|cite|improve this question
























  • you substitute $y=mx+c$ in ellipse equation and make the discriminant of the quadratic zero
    – Ekaveera Kumar Sharma
    Feb 22 '16 at 4:00
















0















If $y=mx + c$ is a tangent to an ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$, then show that $c^2=a^2m^2 + b^2$.




So for this question, first off I tried to differentiate it using implicit differentiation. I got
$$frac{dy}{dx}=-frac{xb^2}{ya^2}$$



Afterward, I was unsure how to proceed. I tried subbing the above into $y=mx +c$, but I don't think that helps. How to I get $c^2=a^2m^2 + b^2$ out?






conic-sections







share|cite|improve this question
























  • you substitute $y=mx+c$ in ellipse equation and make the discriminant of the quadratic zero
    – Ekaveera Kumar Sharma
    Feb 22 '16 at 4:00














0












0








0








If $y=mx + c$ is a tangent to an ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$, then show that $c^2=a^2m^2 + b^2$.




So for this question, first off I tried to differentiate it using implicit differentiation. I got
$$frac{dy}{dx}=-frac{xb^2}{ya^2}$$



Afterward, I was unsure how to proceed. I tried subbing the above into $y=mx +c$, but I don't think that helps. How to I get $c^2=a^2m^2 + b^2$ out?






conic-sections







share|cite|improve this question
















If $y=mx + c$ is a tangent to an ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$, then show that $c^2=a^2m^2 + b^2$.




So for this question, first off I tried to differentiate it using implicit differentiation. I got
$$frac{dy}{dx}=-frac{xb^2}{ya^2}$$



Afterward, I was unsure how to proceed. I tried subbing the above into $y=mx +c$, but I don't think that helps. How to I get $c^2=a^2m^2 + b^2$ out?






conic-sections




conic-sections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 12:28









Blue

47.6k870151




47.6k870151










asked Feb 22 '16 at 3:30









ChemistryStudent

337




337












  • you substitute $y=mx+c$ in ellipse equation and make the discriminant of the quadratic zero
    – Ekaveera Kumar Sharma
    Feb 22 '16 at 4:00


















  • you substitute $y=mx+c$ in ellipse equation and make the discriminant of the quadratic zero
    – Ekaveera Kumar Sharma
    Feb 22 '16 at 4:00
















you substitute $y=mx+c$ in ellipse equation and make the discriminant of the quadratic zero
– Ekaveera Kumar Sharma
Feb 22 '16 at 4:00




you substitute $y=mx+c$ in ellipse equation and make the discriminant of the quadratic zero
– Ekaveera Kumar Sharma
Feb 22 '16 at 4:00










2 Answers
2






active

oldest

votes


















0














Let a general point on the given line be represented by $P=(alpha, malpha+c)$. Then this point's distance taken from the two foci ($(ae,0); (-ae,0)$) must sum to $2a$, where $e=(1-frac{b^2}{a^2})^frac{1}{2}$.
$therefore ((alpha-ae)^2 + (alpha m+c-0)^2))^frac{1}{2} + ((alpha m+ae)^2 + (alpha+c-0)^2))^frac{1}{2} = 2a$
But you don't actually solve need to solve this equation for $alpha$. Get the quadratic equation in $alpha$ and since the line $y=mx+c$ is a tangent, it touches the curve at a single point, hence the determinant of the equation in $alpha$ should be zero. That should give you the result.






share|cite|improve this answer























  • so do I get the quadratic by subbing in y=mx + c into the ellipsis equation?
    – ChemistryStudent
    Feb 22 '16 at 4:08












  • Yes. $(x,mx+c)$ would represent a general point on the line. Get a quadratic in $x$, set the determinant $=0$.
    – Omanshu Thapliyal
    Feb 22 '16 at 4:10



















0














Substitute $y=mx+c$ into $frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1$, we have



begin{align*}
frac{x^{2}}{a^{2}}+frac{(mx+c)^{2}}{b^{2}} &=1 \
left( frac{1}{a^{2}}+frac{m^{2}}{b^{2}} right)x^{2}+
frac{2mc}{b^{2}} x+frac{c^{2}}{b^{2}}-1 &=0
end{align*}



For tangency, $Delta=0$



begin{align*}
left( frac{2mc}{b^{2}} right)^{2}-
4left( frac{1}{a^{2}}+frac{m^{2}}{b^{2}} right)
left( frac{c^{2}}{b^{2}}-1 right) &=0 \
m^{2}a^{2}c^{2}-(b^{2}+a^{2}m^{2})(c^{2}-b^{2}) &= 0 \
b^{4}-b^{2}c^{2}+a^{2}b^{2}m^{2} &=0 \
a^{2}m^{2}+b^{2} &= c^{2}
end{align*}






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%2f1666529%2fif-y-mx-c-is-a-tangent-to-an-ellipse-fracx2a2-fracy2b2-1-th%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    Let a general point on the given line be represented by $P=(alpha, malpha+c)$. Then this point's distance taken from the two foci ($(ae,0); (-ae,0)$) must sum to $2a$, where $e=(1-frac{b^2}{a^2})^frac{1}{2}$.
    $therefore ((alpha-ae)^2 + (alpha m+c-0)^2))^frac{1}{2} + ((alpha m+ae)^2 + (alpha+c-0)^2))^frac{1}{2} = 2a$
    But you don't actually solve need to solve this equation for $alpha$. Get the quadratic equation in $alpha$ and since the line $y=mx+c$ is a tangent, it touches the curve at a single point, hence the determinant of the equation in $alpha$ should be zero. That should give you the result.






    share|cite|improve this answer























    • so do I get the quadratic by subbing in y=mx + c into the ellipsis equation?
      – ChemistryStudent
      Feb 22 '16 at 4:08












    • Yes. $(x,mx+c)$ would represent a general point on the line. Get a quadratic in $x$, set the determinant $=0$.
      – Omanshu Thapliyal
      Feb 22 '16 at 4:10
















    0














    Let a general point on the given line be represented by $P=(alpha, malpha+c)$. Then this point's distance taken from the two foci ($(ae,0); (-ae,0)$) must sum to $2a$, where $e=(1-frac{b^2}{a^2})^frac{1}{2}$.
    $therefore ((alpha-ae)^2 + (alpha m+c-0)^2))^frac{1}{2} + ((alpha m+ae)^2 + (alpha+c-0)^2))^frac{1}{2} = 2a$
    But you don't actually solve need to solve this equation for $alpha$. Get the quadratic equation in $alpha$ and since the line $y=mx+c$ is a tangent, it touches the curve at a single point, hence the determinant of the equation in $alpha$ should be zero. That should give you the result.






    share|cite|improve this answer























    • so do I get the quadratic by subbing in y=mx + c into the ellipsis equation?
      – ChemistryStudent
      Feb 22 '16 at 4:08












    • Yes. $(x,mx+c)$ would represent a general point on the line. Get a quadratic in $x$, set the determinant $=0$.
      – Omanshu Thapliyal
      Feb 22 '16 at 4:10














    0












    0








    0






    Let a general point on the given line be represented by $P=(alpha, malpha+c)$. Then this point's distance taken from the two foci ($(ae,0); (-ae,0)$) must sum to $2a$, where $e=(1-frac{b^2}{a^2})^frac{1}{2}$.
    $therefore ((alpha-ae)^2 + (alpha m+c-0)^2))^frac{1}{2} + ((alpha m+ae)^2 + (alpha+c-0)^2))^frac{1}{2} = 2a$
    But you don't actually solve need to solve this equation for $alpha$. Get the quadratic equation in $alpha$ and since the line $y=mx+c$ is a tangent, it touches the curve at a single point, hence the determinant of the equation in $alpha$ should be zero. That should give you the result.






    share|cite|improve this answer














    Let a general point on the given line be represented by $P=(alpha, malpha+c)$. Then this point's distance taken from the two foci ($(ae,0); (-ae,0)$) must sum to $2a$, where $e=(1-frac{b^2}{a^2})^frac{1}{2}$.
    $therefore ((alpha-ae)^2 + (alpha m+c-0)^2))^frac{1}{2} + ((alpha m+ae)^2 + (alpha+c-0)^2))^frac{1}{2} = 2a$
    But you don't actually solve need to solve this equation for $alpha$. Get the quadratic equation in $alpha$ and since the line $y=mx+c$ is a tangent, it touches the curve at a single point, hence the determinant of the equation in $alpha$ should be zero. That should give you the result.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Feb 22 '16 at 4:45

























    answered Feb 22 '16 at 4:01









    Omanshu Thapliyal

    14




    14












    • so do I get the quadratic by subbing in y=mx + c into the ellipsis equation?
      – ChemistryStudent
      Feb 22 '16 at 4:08












    • Yes. $(x,mx+c)$ would represent a general point on the line. Get a quadratic in $x$, set the determinant $=0$.
      – Omanshu Thapliyal
      Feb 22 '16 at 4:10


















    • so do I get the quadratic by subbing in y=mx + c into the ellipsis equation?
      – ChemistryStudent
      Feb 22 '16 at 4:08












    • Yes. $(x,mx+c)$ would represent a general point on the line. Get a quadratic in $x$, set the determinant $=0$.
      – Omanshu Thapliyal
      Feb 22 '16 at 4:10
















    so do I get the quadratic by subbing in y=mx + c into the ellipsis equation?
    – ChemistryStudent
    Feb 22 '16 at 4:08






    so do I get the quadratic by subbing in y=mx + c into the ellipsis equation?
    – ChemistryStudent
    Feb 22 '16 at 4:08














    Yes. $(x,mx+c)$ would represent a general point on the line. Get a quadratic in $x$, set the determinant $=0$.
    – Omanshu Thapliyal
    Feb 22 '16 at 4:10




    Yes. $(x,mx+c)$ would represent a general point on the line. Get a quadratic in $x$, set the determinant $=0$.
    – Omanshu Thapliyal
    Feb 22 '16 at 4:10











    0














    Substitute $y=mx+c$ into $frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1$, we have



    begin{align*}
    frac{x^{2}}{a^{2}}+frac{(mx+c)^{2}}{b^{2}} &=1 \
    left( frac{1}{a^{2}}+frac{m^{2}}{b^{2}} right)x^{2}+
    frac{2mc}{b^{2}} x+frac{c^{2}}{b^{2}}-1 &=0
    end{align*}



    For tangency, $Delta=0$



    begin{align*}
    left( frac{2mc}{b^{2}} right)^{2}-
    4left( frac{1}{a^{2}}+frac{m^{2}}{b^{2}} right)
    left( frac{c^{2}}{b^{2}}-1 right) &=0 \
    m^{2}a^{2}c^{2}-(b^{2}+a^{2}m^{2})(c^{2}-b^{2}) &= 0 \
    b^{4}-b^{2}c^{2}+a^{2}b^{2}m^{2} &=0 \
    a^{2}m^{2}+b^{2} &= c^{2}
    end{align*}






    share|cite|improve this answer


























      0














      Substitute $y=mx+c$ into $frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1$, we have



      begin{align*}
      frac{x^{2}}{a^{2}}+frac{(mx+c)^{2}}{b^{2}} &=1 \
      left( frac{1}{a^{2}}+frac{m^{2}}{b^{2}} right)x^{2}+
      frac{2mc}{b^{2}} x+frac{c^{2}}{b^{2}}-1 &=0
      end{align*}



      For tangency, $Delta=0$



      begin{align*}
      left( frac{2mc}{b^{2}} right)^{2}-
      4left( frac{1}{a^{2}}+frac{m^{2}}{b^{2}} right)
      left( frac{c^{2}}{b^{2}}-1 right) &=0 \
      m^{2}a^{2}c^{2}-(b^{2}+a^{2}m^{2})(c^{2}-b^{2}) &= 0 \
      b^{4}-b^{2}c^{2}+a^{2}b^{2}m^{2} &=0 \
      a^{2}m^{2}+b^{2} &= c^{2}
      end{align*}






      share|cite|improve this answer
























        0












        0








        0






        Substitute $y=mx+c$ into $frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1$, we have



        begin{align*}
        frac{x^{2}}{a^{2}}+frac{(mx+c)^{2}}{b^{2}} &=1 \
        left( frac{1}{a^{2}}+frac{m^{2}}{b^{2}} right)x^{2}+
        frac{2mc}{b^{2}} x+frac{c^{2}}{b^{2}}-1 &=0
        end{align*}



        For tangency, $Delta=0$



        begin{align*}
        left( frac{2mc}{b^{2}} right)^{2}-
        4left( frac{1}{a^{2}}+frac{m^{2}}{b^{2}} right)
        left( frac{c^{2}}{b^{2}}-1 right) &=0 \
        m^{2}a^{2}c^{2}-(b^{2}+a^{2}m^{2})(c^{2}-b^{2}) &= 0 \
        b^{4}-b^{2}c^{2}+a^{2}b^{2}m^{2} &=0 \
        a^{2}m^{2}+b^{2} &= c^{2}
        end{align*}






        share|cite|improve this answer












        Substitute $y=mx+c$ into $frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1$, we have



        begin{align*}
        frac{x^{2}}{a^{2}}+frac{(mx+c)^{2}}{b^{2}} &=1 \
        left( frac{1}{a^{2}}+frac{m^{2}}{b^{2}} right)x^{2}+
        frac{2mc}{b^{2}} x+frac{c^{2}}{b^{2}}-1 &=0
        end{align*}



        For tangency, $Delta=0$



        begin{align*}
        left( frac{2mc}{b^{2}} right)^{2}-
        4left( frac{1}{a^{2}}+frac{m^{2}}{b^{2}} right)
        left( frac{c^{2}}{b^{2}}-1 right) &=0 \
        m^{2}a^{2}c^{2}-(b^{2}+a^{2}m^{2})(c^{2}-b^{2}) &= 0 \
        b^{4}-b^{2}c^{2}+a^{2}b^{2}m^{2} &=0 \
        a^{2}m^{2}+b^{2} &= c^{2}
        end{align*}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 22 '16 at 6:57









        Ng Chung Tak

        14k31334




        14k31334






























            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%2f1666529%2fif-y-mx-c-is-a-tangent-to-an-ellipse-fracx2a2-fracy2b2-1-th%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

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
            Read more

            “minimization” problem in Euclidean space related to orthonormal basis

            Image
            3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
            Read more