Tangents to parabola $y^{2}=4ax$ meet hyperbola $x^2/a^2-y^2/b^2=1$ at $A$ and $B$. Find the locus of...












1












$begingroup$



If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $dfrac{x^2}{a^2} - dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of tangents at $A$ and $B$.




I know that tangent to parabola is $y = mx + a/m$ ($m$ being the slope), but I am not able to figure out how to take out point of intersections.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
    $endgroup$
    – Blue
    Dec 29 '18 at 23:30










  • $begingroup$
    I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
    $endgroup$
    – Badguy
    Dec 30 '18 at 8:47












  • $begingroup$
    What is "$T$" ?
    $endgroup$
    – Blue
    Dec 30 '18 at 8:50
















1












$begingroup$



If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $dfrac{x^2}{a^2} - dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of tangents at $A$ and $B$.




I know that tangent to parabola is $y = mx + a/m$ ($m$ being the slope), but I am not able to figure out how to take out point of intersections.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
    $endgroup$
    – Blue
    Dec 29 '18 at 23:30










  • $begingroup$
    I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
    $endgroup$
    – Badguy
    Dec 30 '18 at 8:47












  • $begingroup$
    What is "$T$" ?
    $endgroup$
    – Blue
    Dec 30 '18 at 8:50














1












1








1


1



$begingroup$



If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $dfrac{x^2}{a^2} - dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of tangents at $A$ and $B$.




I know that tangent to parabola is $y = mx + a/m$ ($m$ being the slope), but I am not able to figure out how to take out point of intersections.










share|cite|improve this question











$endgroup$





If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $dfrac{x^2}{a^2} - dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of tangents at $A$ and $B$.




I know that tangent to parabola is $y = mx + a/m$ ($m$ being the slope), but I am not able to figure out how to take out point of intersections.







analytic-geometry conic-sections tangent-line






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share|cite|improve this question













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share|cite|improve this question








edited Dec 29 '18 at 23:30









Blue

47.7k870151




47.7k870151










asked Dec 29 '18 at 15:56









BadguyBadguy

63




63








  • 1




    $begingroup$
    Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
    $endgroup$
    – Blue
    Dec 29 '18 at 23:30










  • $begingroup$
    I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
    $endgroup$
    – Badguy
    Dec 30 '18 at 8:47












  • $begingroup$
    What is "$T$" ?
    $endgroup$
    – Blue
    Dec 30 '18 at 8:50














  • 1




    $begingroup$
    Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
    $endgroup$
    – Blue
    Dec 29 '18 at 23:30










  • $begingroup$
    I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
    $endgroup$
    – Badguy
    Dec 30 '18 at 8:47












  • $begingroup$
    What is "$T$" ?
    $endgroup$
    – Blue
    Dec 30 '18 at 8:50








1




1




$begingroup$
Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
$endgroup$
– Blue
Dec 29 '18 at 23:30




$begingroup$
Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
$endgroup$
– Blue
Dec 29 '18 at 23:30












$begingroup$
I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
$endgroup$
– Badguy
Dec 30 '18 at 8:47






$begingroup$
I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
$endgroup$
– Badguy
Dec 30 '18 at 8:47














$begingroup$
What is "$T$" ?
$endgroup$
– Blue
Dec 30 '18 at 8:50




$begingroup$
What is "$T$" ?
$endgroup$
– Blue
Dec 30 '18 at 8:50










2 Answers
2






active

oldest

votes


















0












$begingroup$

Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Not understood.
    $endgroup$
    – Badguy
    Dec 29 '18 at 16:22










  • $begingroup$
    I have neither studied nor seen what youve written here
    $endgroup$
    – Badguy
    Dec 29 '18 at 16:23










  • $begingroup$
    @Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
    $endgroup$
    – Ng Chung Tak
    Jan 2 at 8:46





















0












$begingroup$

$$y^2=4ax tag{1}$$



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



Let $P(X,Y)$ be the required locus.





  • For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).



    Equation of $AB$ is
    $$frac{X x}{a^2}-frac{Y y}{b^2}=1 tag{3}$$




  • Equation of tangent of $(1)$ at $C(x_1,y_1)$
    $$y_1 y=2a(x+x_1)$$



    Rearranging, we have
    $$-frac{x}{x_1}+frac{y_1 y}{2a x_1}=1 tag{4}$$




  • Identifying $(3)$ and $(4)$, we get
    $$(X,Y)=left( -frac{a^2}{x_1}, -frac{b^2 y_1}{2a x_1} right)$$



    $$(x_1,y_1)=left( -frac{a^2}{X}, frac{2a^3 Y}{b^2 X} right)$$



    But $$y_1^2=4a x_1$$



    $$left( frac{2a^3 Y}{b^2 X} right)^2=4aleft( -frac{a^2}{X} right)$$




The locus of $P$ is



$$fbox{$a^3 Y^2+b^4 X=0$}$$



enter image description here




Useful fact:



Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by



$$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$







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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Not understood.
      $endgroup$
      – Badguy
      Dec 29 '18 at 16:22










    • $begingroup$
      I have neither studied nor seen what youve written here
      $endgroup$
      – Badguy
      Dec 29 '18 at 16:23










    • $begingroup$
      @Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
      $endgroup$
      – Ng Chung Tak
      Jan 2 at 8:46


















    0












    $begingroup$

    Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Not understood.
      $endgroup$
      – Badguy
      Dec 29 '18 at 16:22










    • $begingroup$
      I have neither studied nor seen what youve written here
      $endgroup$
      – Badguy
      Dec 29 '18 at 16:23










    • $begingroup$
      @Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
      $endgroup$
      – Ng Chung Tak
      Jan 2 at 8:46
















    0












    0








    0





    $begingroup$

    Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$






    share|cite|improve this answer









    $endgroup$



    Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 29 '18 at 16:12









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    73.5k42865




    73.5k42865












    • $begingroup$
      Not understood.
      $endgroup$
      – Badguy
      Dec 29 '18 at 16:22










    • $begingroup$
      I have neither studied nor seen what youve written here
      $endgroup$
      – Badguy
      Dec 29 '18 at 16:23










    • $begingroup$
      @Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
      $endgroup$
      – Ng Chung Tak
      Jan 2 at 8:46




















    • $begingroup$
      Not understood.
      $endgroup$
      – Badguy
      Dec 29 '18 at 16:22










    • $begingroup$
      I have neither studied nor seen what youve written here
      $endgroup$
      – Badguy
      Dec 29 '18 at 16:23










    • $begingroup$
      @Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
      $endgroup$
      – Ng Chung Tak
      Jan 2 at 8:46


















    $begingroup$
    Not understood.
    $endgroup$
    – Badguy
    Dec 29 '18 at 16:22




    $begingroup$
    Not understood.
    $endgroup$
    – Badguy
    Dec 29 '18 at 16:22












    $begingroup$
    I have neither studied nor seen what youve written here
    $endgroup$
    – Badguy
    Dec 29 '18 at 16:23




    $begingroup$
    I have neither studied nor seen what youve written here
    $endgroup$
    – Badguy
    Dec 29 '18 at 16:23












    $begingroup$
    @Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
    $endgroup$
    – Ng Chung Tak
    Jan 2 at 8:46






    $begingroup$
    @Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
    $endgroup$
    – Ng Chung Tak
    Jan 2 at 8:46













    0












    $begingroup$

    $$y^2=4ax tag{1}$$



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



    Let $P(X,Y)$ be the required locus.





    • For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).



      Equation of $AB$ is
      $$frac{X x}{a^2}-frac{Y y}{b^2}=1 tag{3}$$




    • Equation of tangent of $(1)$ at $C(x_1,y_1)$
      $$y_1 y=2a(x+x_1)$$



      Rearranging, we have
      $$-frac{x}{x_1}+frac{y_1 y}{2a x_1}=1 tag{4}$$




    • Identifying $(3)$ and $(4)$, we get
      $$(X,Y)=left( -frac{a^2}{x_1}, -frac{b^2 y_1}{2a x_1} right)$$



      $$(x_1,y_1)=left( -frac{a^2}{X}, frac{2a^3 Y}{b^2 X} right)$$



      But $$y_1^2=4a x_1$$



      $$left( frac{2a^3 Y}{b^2 X} right)^2=4aleft( -frac{a^2}{X} right)$$




    The locus of $P$ is



    $$fbox{$a^3 Y^2+b^4 X=0$}$$



    enter image description here




    Useful fact:



    Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by



    $$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$







    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      $$y^2=4ax tag{1}$$



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



      Let $P(X,Y)$ be the required locus.





      • For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).



        Equation of $AB$ is
        $$frac{X x}{a^2}-frac{Y y}{b^2}=1 tag{3}$$




      • Equation of tangent of $(1)$ at $C(x_1,y_1)$
        $$y_1 y=2a(x+x_1)$$



        Rearranging, we have
        $$-frac{x}{x_1}+frac{y_1 y}{2a x_1}=1 tag{4}$$




      • Identifying $(3)$ and $(4)$, we get
        $$(X,Y)=left( -frac{a^2}{x_1}, -frac{b^2 y_1}{2a x_1} right)$$



        $$(x_1,y_1)=left( -frac{a^2}{X}, frac{2a^3 Y}{b^2 X} right)$$



        But $$y_1^2=4a x_1$$



        $$left( frac{2a^3 Y}{b^2 X} right)^2=4aleft( -frac{a^2}{X} right)$$




      The locus of $P$ is



      $$fbox{$a^3 Y^2+b^4 X=0$}$$



      enter image description here




      Useful fact:



      Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by



      $$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$







      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        $$y^2=4ax tag{1}$$



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



        Let $P(X,Y)$ be the required locus.





        • For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).



          Equation of $AB$ is
          $$frac{X x}{a^2}-frac{Y y}{b^2}=1 tag{3}$$




        • Equation of tangent of $(1)$ at $C(x_1,y_1)$
          $$y_1 y=2a(x+x_1)$$



          Rearranging, we have
          $$-frac{x}{x_1}+frac{y_1 y}{2a x_1}=1 tag{4}$$




        • Identifying $(3)$ and $(4)$, we get
          $$(X,Y)=left( -frac{a^2}{x_1}, -frac{b^2 y_1}{2a x_1} right)$$



          $$(x_1,y_1)=left( -frac{a^2}{X}, frac{2a^3 Y}{b^2 X} right)$$



          But $$y_1^2=4a x_1$$



          $$left( frac{2a^3 Y}{b^2 X} right)^2=4aleft( -frac{a^2}{X} right)$$




        The locus of $P$ is



        $$fbox{$a^3 Y^2+b^4 X=0$}$$



        enter image description here




        Useful fact:



        Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by



        $$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$







        share|cite|improve this answer











        $endgroup$



        $$y^2=4ax tag{1}$$



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



        Let $P(X,Y)$ be the required locus.





        • For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).



          Equation of $AB$ is
          $$frac{X x}{a^2}-frac{Y y}{b^2}=1 tag{3}$$




        • Equation of tangent of $(1)$ at $C(x_1,y_1)$
          $$y_1 y=2a(x+x_1)$$



          Rearranging, we have
          $$-frac{x}{x_1}+frac{y_1 y}{2a x_1}=1 tag{4}$$




        • Identifying $(3)$ and $(4)$, we get
          $$(X,Y)=left( -frac{a^2}{x_1}, -frac{b^2 y_1}{2a x_1} right)$$



          $$(x_1,y_1)=left( -frac{a^2}{X}, frac{2a^3 Y}{b^2 X} right)$$



          But $$y_1^2=4a x_1$$



          $$left( frac{2a^3 Y}{b^2 X} right)^2=4aleft( -frac{a^2}{X} right)$$




        The locus of $P$ is



        $$fbox{$a^3 Y^2+b^4 X=0$}$$



        enter image description here




        Useful fact:



        Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by



        $$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 9:47

























        answered Jan 2 at 9:37









        Ng Chung TakNg Chung Tak

        14.3k31334




        14.3k31334






























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