Find the locus of points $P$ for which angle bisector of tangents to a standard ellipse is $y = x tan theta$.












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Find the locus of points $P$ for which angle bisector of tangents to a standard ellipse is $y = x tan theta$.




Let $ P = (u, v)$ and ellipse be $E = dfrac{x^2}{a^2} + dfrac{y^2}{b^2} - 1$



Then joint tangent equation for the point $P$ is $$J(x,y) = dfrac{(y-v)^2}{b^2} + dfrac{(x-u)^2}{a^2} - dfrac{(uy-vx)^2}{a^2b^2} $$



I have to find $(u,v)$ for which the angle bisectors of line pair $J$ are $ y - xtan theta = 0$ and $y + xcot theta -v - u cot theta = 0$.



Translating $P$ - the centre of $J$ - to origin with $x = X + u$ and $y = Y + v$:



$$J'(X, Y) = J(X+u, Y+u) = X^2 left (dfrac1{a^2} - dfrac{v^2}{a^2b^2}right)
+ Y^2 left (dfrac1{b^2} - dfrac{u^2}{a^2b^2}right) + dfrac{2XYuv}{a^2 b^2}$$



The translated angle bisectors are $Y - Xtan theta + v - u tan theta = 0$ and $Y + X cot theta = 0$.



The angle bisectors of $J'$ is given by $$X^2 dfrac{uv}{a^2b^2} + left(dfrac1 {b^2} -dfrac{u^2}{a^2b^2} -dfrac{1}{a^2} +dfrac{v^2}{a^2b^2} right)XY - dfrac{uv}{a^2b^2} Y^2 = 0tag 1$$



or $$(Y + X cot theta) (Y - tan theta X + v - u tan theta) = 0tag 2$$



Equating corresponding terms of $(1)$ and $(2)$ gives $$begin{cases}3) : uv = -a^2b^2 \4) : v = u tan theta \5) : dfrac1 {b^2} -dfrac{u^2}{a^2b^2} -dfrac{1}{a^2} +dfrac{v^2}{a^2b^2} = cot theta - tan theta end{cases}$$



From $(3)$ and $(5)$ I get $u^2 - v^2 -uv(cot theta - tan theta) = a^2 - b^2$




The answer given is $u^2 - v^2 -2uvcot theta = a^2 - b^2$.



Where did I go wrong ?



Is there any alternative way to do this questions, my method is very tedious ?






geometry conic-sections







share|cite|improve this question











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    1












    $begingroup$



    Find the locus of points $P$ for which angle bisector of tangents to a standard ellipse is $y = x tan theta$.




    Let $ P = (u, v)$ and ellipse be $E = dfrac{x^2}{a^2} + dfrac{y^2}{b^2} - 1$



    Then joint tangent equation for the point $P$ is $$J(x,y) = dfrac{(y-v)^2}{b^2} + dfrac{(x-u)^2}{a^2} - dfrac{(uy-vx)^2}{a^2b^2} $$



    I have to find $(u,v)$ for which the angle bisectors of line pair $J$ are $ y - xtan theta = 0$ and $y + xcot theta -v - u cot theta = 0$.



    Translating $P$ - the centre of $J$ - to origin with $x = X + u$ and $y = Y + v$:



    $$J'(X, Y) = J(X+u, Y+u) = X^2 left (dfrac1{a^2} - dfrac{v^2}{a^2b^2}right)
    + Y^2 left (dfrac1{b^2} - dfrac{u^2}{a^2b^2}right) + dfrac{2XYuv}{a^2 b^2}$$



    The translated angle bisectors are $Y - Xtan theta + v - u tan theta = 0$ and $Y + X cot theta = 0$.



    The angle bisectors of $J'$ is given by $$X^2 dfrac{uv}{a^2b^2} + left(dfrac1 {b^2} -dfrac{u^2}{a^2b^2} -dfrac{1}{a^2} +dfrac{v^2}{a^2b^2} right)XY - dfrac{uv}{a^2b^2} Y^2 = 0tag 1$$



    or $$(Y + X cot theta) (Y - tan theta X + v - u tan theta) = 0tag 2$$



    Equating corresponding terms of $(1)$ and $(2)$ gives $$begin{cases}3) : uv = -a^2b^2 \4) : v = u tan theta \5) : dfrac1 {b^2} -dfrac{u^2}{a^2b^2} -dfrac{1}{a^2} +dfrac{v^2}{a^2b^2} = cot theta - tan theta end{cases}$$



    From $(3)$ and $(5)$ I get $u^2 - v^2 -uv(cot theta - tan theta) = a^2 - b^2$




    The answer given is $u^2 - v^2 -2uvcot theta = a^2 - b^2$.



    Where did I go wrong ?



    Is there any alternative way to do this questions, my method is very tedious ?






    geometry conic-sections







    share|cite|improve this question











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      1








      1


      0



      $begingroup$



      Find the locus of points $P$ for which angle bisector of tangents to a standard ellipse is $y = x tan theta$.




      Let $ P = (u, v)$ and ellipse be $E = dfrac{x^2}{a^2} + dfrac{y^2}{b^2} - 1$



      Then joint tangent equation for the point $P$ is $$J(x,y) = dfrac{(y-v)^2}{b^2} + dfrac{(x-u)^2}{a^2} - dfrac{(uy-vx)^2}{a^2b^2} $$



      I have to find $(u,v)$ for which the angle bisectors of line pair $J$ are $ y - xtan theta = 0$ and $y + xcot theta -v - u cot theta = 0$.



      Translating $P$ - the centre of $J$ - to origin with $x = X + u$ and $y = Y + v$:



      $$J'(X, Y) = J(X+u, Y+u) = X^2 left (dfrac1{a^2} - dfrac{v^2}{a^2b^2}right)
      + Y^2 left (dfrac1{b^2} - dfrac{u^2}{a^2b^2}right) + dfrac{2XYuv}{a^2 b^2}$$



      The translated angle bisectors are $Y - Xtan theta + v - u tan theta = 0$ and $Y + X cot theta = 0$.



      The angle bisectors of $J'$ is given by $$X^2 dfrac{uv}{a^2b^2} + left(dfrac1 {b^2} -dfrac{u^2}{a^2b^2} -dfrac{1}{a^2} +dfrac{v^2}{a^2b^2} right)XY - dfrac{uv}{a^2b^2} Y^2 = 0tag 1$$



      or $$(Y + X cot theta) (Y - tan theta X + v - u tan theta) = 0tag 2$$



      Equating corresponding terms of $(1)$ and $(2)$ gives $$begin{cases}3) : uv = -a^2b^2 \4) : v = u tan theta \5) : dfrac1 {b^2} -dfrac{u^2}{a^2b^2} -dfrac{1}{a^2} +dfrac{v^2}{a^2b^2} = cot theta - tan theta end{cases}$$



      From $(3)$ and $(5)$ I get $u^2 - v^2 -uv(cot theta - tan theta) = a^2 - b^2$




      The answer given is $u^2 - v^2 -2uvcot theta = a^2 - b^2$.



      Where did I go wrong ?



      Is there any alternative way to do this questions, my method is very tedious ?






      geometry conic-sections







      share|cite|improve this question











      $endgroup$





      Find the locus of points $P$ for which angle bisector of tangents to a standard ellipse is $y = x tan theta$.




      Let $ P = (u, v)$ and ellipse be $E = dfrac{x^2}{a^2} + dfrac{y^2}{b^2} - 1$



      Then joint tangent equation for the point $P$ is $$J(x,y) = dfrac{(y-v)^2}{b^2} + dfrac{(x-u)^2}{a^2} - dfrac{(uy-vx)^2}{a^2b^2} $$



      I have to find $(u,v)$ for which the angle bisectors of line pair $J$ are $ y - xtan theta = 0$ and $y + xcot theta -v - u cot theta = 0$.



      Translating $P$ - the centre of $J$ - to origin with $x = X + u$ and $y = Y + v$:



      $$J'(X, Y) = J(X+u, Y+u) = X^2 left (dfrac1{a^2} - dfrac{v^2}{a^2b^2}right)
      + Y^2 left (dfrac1{b^2} - dfrac{u^2}{a^2b^2}right) + dfrac{2XYuv}{a^2 b^2}$$



      The translated angle bisectors are $Y - Xtan theta + v - u tan theta = 0$ and $Y + X cot theta = 0$.



      The angle bisectors of $J'$ is given by $$X^2 dfrac{uv}{a^2b^2} + left(dfrac1 {b^2} -dfrac{u^2}{a^2b^2} -dfrac{1}{a^2} +dfrac{v^2}{a^2b^2} right)XY - dfrac{uv}{a^2b^2} Y^2 = 0tag 1$$



      or $$(Y + X cot theta) (Y - tan theta X + v - u tan theta) = 0tag 2$$



      Equating corresponding terms of $(1)$ and $(2)$ gives $$begin{cases}3) : uv = -a^2b^2 \4) : v = u tan theta \5) : dfrac1 {b^2} -dfrac{u^2}{a^2b^2} -dfrac{1}{a^2} +dfrac{v^2}{a^2b^2} = cot theta - tan theta end{cases}$$



      From $(3)$ and $(5)$ I get $u^2 - v^2 -uv(cot theta - tan theta) = a^2 - b^2$




      The answer given is $u^2 - v^2 -2uvcot theta = a^2 - b^2$.



      Where did I go wrong ?



      Is there any alternative way to do this questions, my method is very tedious ?






      geometry conic-sections




      geometry conic-sections






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      edited Feb 3 at 11:25









      Blue

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      asked Feb 3 at 11:24









      user8277998user8277998

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          I think there is something wrong in this question, unless I'm missing something. Requesting that the angle bisector of the tangents issued from $P$ is $y=xtantheta$ is the same as requesting that the angle bisector passes through $O=(0,0)$, that is through the center of the ellipse (assuming that a "standard ellipse" has its center there).



          If $A$ and $B$ are the tangency points and $M$ their midpoint, it is a property of the ellipse that line $PM$ passes through $O$. Hence, to satisfy the request, the bisector of $angle APB$ must pass through $M$, which entails $PA=PB$. But this condition is satisfied only if $P$ belongs to the axes of the ellipse (unless the ellipse is a circle).



          In addition, I don't understand the role of $theta$: if it is given (as it seems to be), then $P$ must belong to the given line $y=xtantheta$, what is the meaning of asking for the locus of $P$?



          EDIT.



          Your solution is right, but substituting there $tantheta=v/u$ (from eq. $(4)$) you get $0=a^2-b^2$. The only way to have a solution is then if $a=b$ (the ellipse is a circle) or if either $tantheta$ or $cottheta$ doesn't exist ($P$ lies on the coordinate axes).






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            $begingroup$

            I think there is something wrong in this question, unless I'm missing something. Requesting that the angle bisector of the tangents issued from $P$ is $y=xtantheta$ is the same as requesting that the angle bisector passes through $O=(0,0)$, that is through the center of the ellipse (assuming that a "standard ellipse" has its center there).



            If $A$ and $B$ are the tangency points and $M$ their midpoint, it is a property of the ellipse that line $PM$ passes through $O$. Hence, to satisfy the request, the bisector of $angle APB$ must pass through $M$, which entails $PA=PB$. But this condition is satisfied only if $P$ belongs to the axes of the ellipse (unless the ellipse is a circle).



            In addition, I don't understand the role of $theta$: if it is given (as it seems to be), then $P$ must belong to the given line $y=xtantheta$, what is the meaning of asking for the locus of $P$?



            EDIT.



            Your solution is right, but substituting there $tantheta=v/u$ (from eq. $(4)$) you get $0=a^2-b^2$. The only way to have a solution is then if $a=b$ (the ellipse is a circle) or if either $tantheta$ or $cottheta$ doesn't exist ($P$ lies on the coordinate axes).






            share|cite|improve this answer











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              $begingroup$

              I think there is something wrong in this question, unless I'm missing something. Requesting that the angle bisector of the tangents issued from $P$ is $y=xtantheta$ is the same as requesting that the angle bisector passes through $O=(0,0)$, that is through the center of the ellipse (assuming that a "standard ellipse" has its center there).



              If $A$ and $B$ are the tangency points and $M$ their midpoint, it is a property of the ellipse that line $PM$ passes through $O$. Hence, to satisfy the request, the bisector of $angle APB$ must pass through $M$, which entails $PA=PB$. But this condition is satisfied only if $P$ belongs to the axes of the ellipse (unless the ellipse is a circle).



              In addition, I don't understand the role of $theta$: if it is given (as it seems to be), then $P$ must belong to the given line $y=xtantheta$, what is the meaning of asking for the locus of $P$?



              EDIT.



              Your solution is right, but substituting there $tantheta=v/u$ (from eq. $(4)$) you get $0=a^2-b^2$. The only way to have a solution is then if $a=b$ (the ellipse is a circle) or if either $tantheta$ or $cottheta$ doesn't exist ($P$ lies on the coordinate axes).






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                I think there is something wrong in this question, unless I'm missing something. Requesting that the angle bisector of the tangents issued from $P$ is $y=xtantheta$ is the same as requesting that the angle bisector passes through $O=(0,0)$, that is through the center of the ellipse (assuming that a "standard ellipse" has its center there).



                If $A$ and $B$ are the tangency points and $M$ their midpoint, it is a property of the ellipse that line $PM$ passes through $O$. Hence, to satisfy the request, the bisector of $angle APB$ must pass through $M$, which entails $PA=PB$. But this condition is satisfied only if $P$ belongs to the axes of the ellipse (unless the ellipse is a circle).



                In addition, I don't understand the role of $theta$: if it is given (as it seems to be), then $P$ must belong to the given line $y=xtantheta$, what is the meaning of asking for the locus of $P$?



                EDIT.



                Your solution is right, but substituting there $tantheta=v/u$ (from eq. $(4)$) you get $0=a^2-b^2$. The only way to have a solution is then if $a=b$ (the ellipse is a circle) or if either $tantheta$ or $cottheta$ doesn't exist ($P$ lies on the coordinate axes).






                share|cite|improve this answer











                $endgroup$



                I think there is something wrong in this question, unless I'm missing something. Requesting that the angle bisector of the tangents issued from $P$ is $y=xtantheta$ is the same as requesting that the angle bisector passes through $O=(0,0)$, that is through the center of the ellipse (assuming that a "standard ellipse" has its center there).



                If $A$ and $B$ are the tangency points and $M$ their midpoint, it is a property of the ellipse that line $PM$ passes through $O$. Hence, to satisfy the request, the bisector of $angle APB$ must pass through $M$, which entails $PA=PB$. But this condition is satisfied only if $P$ belongs to the axes of the ellipse (unless the ellipse is a circle).



                In addition, I don't understand the role of $theta$: if it is given (as it seems to be), then $P$ must belong to the given line $y=xtantheta$, what is the meaning of asking for the locus of $P$?



                EDIT.



                Your solution is right, but substituting there $tantheta=v/u$ (from eq. $(4)$) you get $0=a^2-b^2$. The only way to have a solution is then if $a=b$ (the ellipse is a circle) or if either $tantheta$ or $cottheta$ doesn't exist ($P$ lies on the coordinate axes).







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 3 at 18:24

























                answered Feb 3 at 14:31









                AretinoAretino

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