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











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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?






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  • 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















up vote
0
down vote

favorite













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













up vote
0
down vote

favorite









up vote
0
down vote

favorite












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






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share|cite|improve this question













share|cite|improve this question




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edited Nov 27 at 12:28









Blue

47.2k870149




47.2k870149










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






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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


















up vote
0
down vote













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*}






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    2 Answers
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    2 Answers
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    up vote
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    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















    up vote
    0
    down vote













    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













    up vote
    0
    down vote










    up vote
    0
    down vote









    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










    up vote
    0
    down vote













    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

























      up vote
      0
      down vote













      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























        up vote
        0
        down vote










        up vote
        0
        down vote









        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

        13.8k31234




        13.8k31234






























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