The equation of the tangent to conic $x^2 - y^2 - 8x + 2y + 11 = 0$ at $(2,1)$ is?











up vote
-2
down vote

favorite












Options are :-



A) $x+2=0$
B) $2x+1=0$
C) $x-2=0$
D) $x+y+1=0$



I have tried using differentiation , find out the slope But, its 0. So, I'm not getting my answer correct.!!?






conic-sections







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  • Please show us how you reached this conclusion.
    – Yves Daoust
    Nov 28 at 8:48










  • Are you sure it’s the numerator of the slope that’s zero?
    – amd
    Nov 28 at 8:50










  • $$(x-2)(2cdot2-8)+(y-1)(-2cdot1+2)=0$$
    – Yves Daoust
    Nov 28 at 8:59












  • Use math.stackexchange.com/questions/774250/… OR math.stackexchange.com/questions/2254073/… or math.stackexchange.com/questions/330593/…
    – lab bhattacharjee
    Nov 28 at 9:06










  • Also: math.stackexchange.com/questions/3009978/…
    – lab bhattacharjee
    Nov 28 at 11:27















up vote
-2
down vote

favorite












Options are :-



A) $x+2=0$
B) $2x+1=0$
C) $x-2=0$
D) $x+y+1=0$



I have tried using differentiation , find out the slope But, its 0. So, I'm not getting my answer correct.!!?






conic-sections







share|cite|improve this question






















  • Please show us how you reached this conclusion.
    – Yves Daoust
    Nov 28 at 8:48










  • Are you sure it’s the numerator of the slope that’s zero?
    – amd
    Nov 28 at 8:50










  • $$(x-2)(2cdot2-8)+(y-1)(-2cdot1+2)=0$$
    – Yves Daoust
    Nov 28 at 8:59












  • Use math.stackexchange.com/questions/774250/… OR math.stackexchange.com/questions/2254073/… or math.stackexchange.com/questions/330593/…
    – lab bhattacharjee
    Nov 28 at 9:06










  • Also: math.stackexchange.com/questions/3009978/…
    – lab bhattacharjee
    Nov 28 at 11:27













up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











Options are :-



A) $x+2=0$
B) $2x+1=0$
C) $x-2=0$
D) $x+y+1=0$



I have tried using differentiation , find out the slope But, its 0. So, I'm not getting my answer correct.!!?






conic-sections







share|cite|improve this question













Options are :-



A) $x+2=0$
B) $2x+1=0$
C) $x-2=0$
D) $x+y+1=0$



I have tried using differentiation , find out the slope But, its 0. So, I'm not getting my answer correct.!!?






conic-sections




conic-sections






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Nov 28 at 8:45









Alvin Carter

13




13












  • Please show us how you reached this conclusion.
    – Yves Daoust
    Nov 28 at 8:48










  • Are you sure it’s the numerator of the slope that’s zero?
    – amd
    Nov 28 at 8:50










  • $$(x-2)(2cdot2-8)+(y-1)(-2cdot1+2)=0$$
    – Yves Daoust
    Nov 28 at 8:59












  • Use math.stackexchange.com/questions/774250/… OR math.stackexchange.com/questions/2254073/… or math.stackexchange.com/questions/330593/…
    – lab bhattacharjee
    Nov 28 at 9:06










  • Also: math.stackexchange.com/questions/3009978/…
    – lab bhattacharjee
    Nov 28 at 11:27


















  • Please show us how you reached this conclusion.
    – Yves Daoust
    Nov 28 at 8:48










  • Are you sure it’s the numerator of the slope that’s zero?
    – amd
    Nov 28 at 8:50










  • $$(x-2)(2cdot2-8)+(y-1)(-2cdot1+2)=0$$
    – Yves Daoust
    Nov 28 at 8:59












  • Use math.stackexchange.com/questions/774250/… OR math.stackexchange.com/questions/2254073/… or math.stackexchange.com/questions/330593/…
    – lab bhattacharjee
    Nov 28 at 9:06










  • Also: math.stackexchange.com/questions/3009978/…
    – lab bhattacharjee
    Nov 28 at 11:27
















Please show us how you reached this conclusion.
– Yves Daoust
Nov 28 at 8:48




Please show us how you reached this conclusion.
– Yves Daoust
Nov 28 at 8:48












Are you sure it’s the numerator of the slope that’s zero?
– amd
Nov 28 at 8:50




Are you sure it’s the numerator of the slope that’s zero?
– amd
Nov 28 at 8:50












$$(x-2)(2cdot2-8)+(y-1)(-2cdot1+2)=0$$
– Yves Daoust
Nov 28 at 8:59






$$(x-2)(2cdot2-8)+(y-1)(-2cdot1+2)=0$$
– Yves Daoust
Nov 28 at 8:59














Use math.stackexchange.com/questions/774250/… OR math.stackexchange.com/questions/2254073/… or math.stackexchange.com/questions/330593/…
– lab bhattacharjee
Nov 28 at 9:06




Use math.stackexchange.com/questions/774250/… OR math.stackexchange.com/questions/2254073/… or math.stackexchange.com/questions/330593/…
– lab bhattacharjee
Nov 28 at 9:06












Also: math.stackexchange.com/questions/3009978/…
– lab bhattacharjee
Nov 28 at 11:27




Also: math.stackexchange.com/questions/3009978/…
– lab bhattacharjee
Nov 28 at 11:27










4 Answers
4






active

oldest

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up vote
1
down vote













Tangent passes through $ (2,1).$



Hence the only option left: $x-2=0.$



Check:



$2x-2y(dy/dx )-8 +2dy/dx=0.$



For $y not =1:$



$dy/dx = dfrac{9-2x}{2(1-y)}$.



What happens for $y rightarrow 1$ ?



(Recall $y rightarrow 1$ implies $x rightarrow 2$)



http://m.wolframalpha.com/input/?i=x%5E2-y%5E2-8x%2B2y%2B11%3D0






share|cite|improve this answer























  • Can you upvote my answer please @Peter? Will be kind of you
    – Akash Roy
    Nov 28 at 9:49










  • Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
    – Akash Roy
    Nov 28 at 9:51


















up vote
0
down vote













You do implicit differentiation to obtain the slope to the tangent. Namely,



$ 2x - 2y dfrac{dy}{dx} - 8 + 2 dfrac{dy}{dx} = 0 $.



Solve for $ dfrac{dy}{dx} $.



Substitute in $ (2, 1) $. This obtains the slope.



Now, it is trivial to find the linear eqn of the line.






share|cite|improve this answer




























    up vote
    0
    down vote













    Clearly, the coefficient of $xy$ in the equation of the given conic is $0$. Therefore, the conic is symmetric about a line that is parallel to the $x$-axis (i.e. $y=1$), and it is symmetric about a line that is parallel to the $y$-axis (i.e. $x=4$).



    Therefore, the equation of the tangent line to the given conic at $(2,1)$ is parallel to the $y$-axis, so it has the equation $x=2$ or $x-2=0$, (which is option C).






    share|cite|improve this answer





















    • can you vote my answer please @ Hussain , will be grateful of you
      – Akash Roy
      Nov 28 at 9:48












    • Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
      – Akash Roy
      Nov 28 at 9:52


















    up vote
    -1
    down vote













    The equation of the tangent should be



    $xx'- yy' -4(x+x') +y+y' +11=0$



    Actually while finding equation of tangent of such curve, an efficient approach is to replace $x^2$ by $xx'$ , $y^2$ by $yy'$ , $x$ by $frac{x+x'}{2}$ and similar for $y$.
    Secondly , replace the values of $x'$ and $y'$ by the coordinates where the tangent is to be found.



    So, substitute $x'=2$ and $y'=1$



    Your answer is option $C$






    share|cite|improve this answer























    • I would kindly ask the down voter what is wrong
      – Akash Roy
      Nov 28 at 8:53










    • Akash.Do not worry about down votes.But: Substitute x=2, y=1......,you can say 1 or 2 words about dy/dx, agree?
      – Peter Szilas
      Nov 28 at 9:58










    • Yeah @Peter , should I explain why the approach works or something like that?
      – Akash Roy
      Nov 28 at 10:05










    • Akash.Your approach is fine.But solving for dy/dx you need to divide by 2(1-y) not =0, division by zero is not defined. And we want y=1! Solve for dy/dx, point out that for y=1, dy/dx is not defined, or something like that(my answer).Is this ok?
      – Peter Szilas
      Nov 28 at 10:23











    Your Answer





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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    Tangent passes through $ (2,1).$



    Hence the only option left: $x-2=0.$



    Check:



    $2x-2y(dy/dx )-8 +2dy/dx=0.$



    For $y not =1:$



    $dy/dx = dfrac{9-2x}{2(1-y)}$.



    What happens for $y rightarrow 1$ ?



    (Recall $y rightarrow 1$ implies $x rightarrow 2$)



    http://m.wolframalpha.com/input/?i=x%5E2-y%5E2-8x%2B2y%2B11%3D0






    share|cite|improve this answer























    • Can you upvote my answer please @Peter? Will be kind of you
      – Akash Roy
      Nov 28 at 9:49










    • Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
      – Akash Roy
      Nov 28 at 9:51















    up vote
    1
    down vote













    Tangent passes through $ (2,1).$



    Hence the only option left: $x-2=0.$



    Check:



    $2x-2y(dy/dx )-8 +2dy/dx=0.$



    For $y not =1:$



    $dy/dx = dfrac{9-2x}{2(1-y)}$.



    What happens for $y rightarrow 1$ ?



    (Recall $y rightarrow 1$ implies $x rightarrow 2$)



    http://m.wolframalpha.com/input/?i=x%5E2-y%5E2-8x%2B2y%2B11%3D0






    share|cite|improve this answer























    • Can you upvote my answer please @Peter? Will be kind of you
      – Akash Roy
      Nov 28 at 9:49










    • Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
      – Akash Roy
      Nov 28 at 9:51













    up vote
    1
    down vote










    up vote
    1
    down vote









    Tangent passes through $ (2,1).$



    Hence the only option left: $x-2=0.$



    Check:



    $2x-2y(dy/dx )-8 +2dy/dx=0.$



    For $y not =1:$



    $dy/dx = dfrac{9-2x}{2(1-y)}$.



    What happens for $y rightarrow 1$ ?



    (Recall $y rightarrow 1$ implies $x rightarrow 2$)



    http://m.wolframalpha.com/input/?i=x%5E2-y%5E2-8x%2B2y%2B11%3D0






    share|cite|improve this answer














    Tangent passes through $ (2,1).$



    Hence the only option left: $x-2=0.$



    Check:



    $2x-2y(dy/dx )-8 +2dy/dx=0.$



    For $y not =1:$



    $dy/dx = dfrac{9-2x}{2(1-y)}$.



    What happens for $y rightarrow 1$ ?



    (Recall $y rightarrow 1$ implies $x rightarrow 2$)



    http://m.wolframalpha.com/input/?i=x%5E2-y%5E2-8x%2B2y%2B11%3D0







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 28 at 9:51

























    answered Nov 28 at 9:29









    Peter Szilas

    10.6k2720




    10.6k2720












    • Can you upvote my answer please @Peter? Will be kind of you
      – Akash Roy
      Nov 28 at 9:49










    • Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
      – Akash Roy
      Nov 28 at 9:51


















    • Can you upvote my answer please @Peter? Will be kind of you
      – Akash Roy
      Nov 28 at 9:49










    • Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
      – Akash Roy
      Nov 28 at 9:51
















    Can you upvote my answer please @Peter? Will be kind of you
    – Akash Roy
    Nov 28 at 9:49




    Can you upvote my answer please @Peter? Will be kind of you
    – Akash Roy
    Nov 28 at 9:49












    Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
    – Akash Roy
    Nov 28 at 9:51




    Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
    – Akash Roy
    Nov 28 at 9:51










    up vote
    0
    down vote













    You do implicit differentiation to obtain the slope to the tangent. Namely,



    $ 2x - 2y dfrac{dy}{dx} - 8 + 2 dfrac{dy}{dx} = 0 $.



    Solve for $ dfrac{dy}{dx} $.



    Substitute in $ (2, 1) $. This obtains the slope.



    Now, it is trivial to find the linear eqn of the line.






    share|cite|improve this answer

























      up vote
      0
      down vote













      You do implicit differentiation to obtain the slope to the tangent. Namely,



      $ 2x - 2y dfrac{dy}{dx} - 8 + 2 dfrac{dy}{dx} = 0 $.



      Solve for $ dfrac{dy}{dx} $.



      Substitute in $ (2, 1) $. This obtains the slope.



      Now, it is trivial to find the linear eqn of the line.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        You do implicit differentiation to obtain the slope to the tangent. Namely,



        $ 2x - 2y dfrac{dy}{dx} - 8 + 2 dfrac{dy}{dx} = 0 $.



        Solve for $ dfrac{dy}{dx} $.



        Substitute in $ (2, 1) $. This obtains the slope.



        Now, it is trivial to find the linear eqn of the line.






        share|cite|improve this answer












        You do implicit differentiation to obtain the slope to the tangent. Namely,



        $ 2x - 2y dfrac{dy}{dx} - 8 + 2 dfrac{dy}{dx} = 0 $.



        Solve for $ dfrac{dy}{dx} $.



        Substitute in $ (2, 1) $. This obtains the slope.



        Now, it is trivial to find the linear eqn of the line.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 8:54









        hhp2122

        163




        163






















            up vote
            0
            down vote













            Clearly, the coefficient of $xy$ in the equation of the given conic is $0$. Therefore, the conic is symmetric about a line that is parallel to the $x$-axis (i.e. $y=1$), and it is symmetric about a line that is parallel to the $y$-axis (i.e. $x=4$).



            Therefore, the equation of the tangent line to the given conic at $(2,1)$ is parallel to the $y$-axis, so it has the equation $x=2$ or $x-2=0$, (which is option C).






            share|cite|improve this answer





















            • can you vote my answer please @ Hussain , will be grateful of you
              – Akash Roy
              Nov 28 at 9:48












            • Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
              – Akash Roy
              Nov 28 at 9:52















            up vote
            0
            down vote













            Clearly, the coefficient of $xy$ in the equation of the given conic is $0$. Therefore, the conic is symmetric about a line that is parallel to the $x$-axis (i.e. $y=1$), and it is symmetric about a line that is parallel to the $y$-axis (i.e. $x=4$).



            Therefore, the equation of the tangent line to the given conic at $(2,1)$ is parallel to the $y$-axis, so it has the equation $x=2$ or $x-2=0$, (which is option C).






            share|cite|improve this answer





















            • can you vote my answer please @ Hussain , will be grateful of you
              – Akash Roy
              Nov 28 at 9:48












            • Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
              – Akash Roy
              Nov 28 at 9:52













            up vote
            0
            down vote










            up vote
            0
            down vote









            Clearly, the coefficient of $xy$ in the equation of the given conic is $0$. Therefore, the conic is symmetric about a line that is parallel to the $x$-axis (i.e. $y=1$), and it is symmetric about a line that is parallel to the $y$-axis (i.e. $x=4$).



            Therefore, the equation of the tangent line to the given conic at $(2,1)$ is parallel to the $y$-axis, so it has the equation $x=2$ or $x-2=0$, (which is option C).






            share|cite|improve this answer












            Clearly, the coefficient of $xy$ in the equation of the given conic is $0$. Therefore, the conic is symmetric about a line that is parallel to the $x$-axis (i.e. $y=1$), and it is symmetric about a line that is parallel to the $y$-axis (i.e. $x=4$).



            Therefore, the equation of the tangent line to the given conic at $(2,1)$ is parallel to the $y$-axis, so it has the equation $x=2$ or $x-2=0$, (which is option C).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 28 at 9:13









            Hussain-Alqatari

            2977




            2977












            • can you vote my answer please @ Hussain , will be grateful of you
              – Akash Roy
              Nov 28 at 9:48












            • Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
              – Akash Roy
              Nov 28 at 9:52


















            • can you vote my answer please @ Hussain , will be grateful of you
              – Akash Roy
              Nov 28 at 9:48












            • Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
              – Akash Roy
              Nov 28 at 9:52
















            can you vote my answer please @ Hussain , will be grateful of you
            – Akash Roy
            Nov 28 at 9:48






            can you vote my answer please @ Hussain , will be grateful of you
            – Akash Roy
            Nov 28 at 9:48














            Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
            – Akash Roy
            Nov 28 at 9:52




            Actually I can't ask questions on this site , however if my positive contributions are good enough to outdo those questions of mine which were poorly received , then the ban would be removed. That's why I am asking for votes. Please cooperate
            – Akash Roy
            Nov 28 at 9:52










            up vote
            -1
            down vote













            The equation of the tangent should be



            $xx'- yy' -4(x+x') +y+y' +11=0$



            Actually while finding equation of tangent of such curve, an efficient approach is to replace $x^2$ by $xx'$ , $y^2$ by $yy'$ , $x$ by $frac{x+x'}{2}$ and similar for $y$.
            Secondly , replace the values of $x'$ and $y'$ by the coordinates where the tangent is to be found.



            So, substitute $x'=2$ and $y'=1$



            Your answer is option $C$






            share|cite|improve this answer























            • I would kindly ask the down voter what is wrong
              – Akash Roy
              Nov 28 at 8:53










            • Akash.Do not worry about down votes.But: Substitute x=2, y=1......,you can say 1 or 2 words about dy/dx, agree?
              – Peter Szilas
              Nov 28 at 9:58










            • Yeah @Peter , should I explain why the approach works or something like that?
              – Akash Roy
              Nov 28 at 10:05










            • Akash.Your approach is fine.But solving for dy/dx you need to divide by 2(1-y) not =0, division by zero is not defined. And we want y=1! Solve for dy/dx, point out that for y=1, dy/dx is not defined, or something like that(my answer).Is this ok?
              – Peter Szilas
              Nov 28 at 10:23















            up vote
            -1
            down vote













            The equation of the tangent should be



            $xx'- yy' -4(x+x') +y+y' +11=0$



            Actually while finding equation of tangent of such curve, an efficient approach is to replace $x^2$ by $xx'$ , $y^2$ by $yy'$ , $x$ by $frac{x+x'}{2}$ and similar for $y$.
            Secondly , replace the values of $x'$ and $y'$ by the coordinates where the tangent is to be found.



            So, substitute $x'=2$ and $y'=1$



            Your answer is option $C$






            share|cite|improve this answer























            • I would kindly ask the down voter what is wrong
              – Akash Roy
              Nov 28 at 8:53










            • Akash.Do not worry about down votes.But: Substitute x=2, y=1......,you can say 1 or 2 words about dy/dx, agree?
              – Peter Szilas
              Nov 28 at 9:58










            • Yeah @Peter , should I explain why the approach works or something like that?
              – Akash Roy
              Nov 28 at 10:05










            • Akash.Your approach is fine.But solving for dy/dx you need to divide by 2(1-y) not =0, division by zero is not defined. And we want y=1! Solve for dy/dx, point out that for y=1, dy/dx is not defined, or something like that(my answer).Is this ok?
              – Peter Szilas
              Nov 28 at 10:23













            up vote
            -1
            down vote










            up vote
            -1
            down vote









            The equation of the tangent should be



            $xx'- yy' -4(x+x') +y+y' +11=0$



            Actually while finding equation of tangent of such curve, an efficient approach is to replace $x^2$ by $xx'$ , $y^2$ by $yy'$ , $x$ by $frac{x+x'}{2}$ and similar for $y$.
            Secondly , replace the values of $x'$ and $y'$ by the coordinates where the tangent is to be found.



            So, substitute $x'=2$ and $y'=1$



            Your answer is option $C$






            share|cite|improve this answer














            The equation of the tangent should be



            $xx'- yy' -4(x+x') +y+y' +11=0$



            Actually while finding equation of tangent of such curve, an efficient approach is to replace $x^2$ by $xx'$ , $y^2$ by $yy'$ , $x$ by $frac{x+x'}{2}$ and similar for $y$.
            Secondly , replace the values of $x'$ and $y'$ by the coordinates where the tangent is to be found.



            So, substitute $x'=2$ and $y'=1$



            Your answer is option $C$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 28 at 9:46

























            answered Nov 28 at 8:51









            Akash Roy

            1




            1












            • I would kindly ask the down voter what is wrong
              – Akash Roy
              Nov 28 at 8:53










            • Akash.Do not worry about down votes.But: Substitute x=2, y=1......,you can say 1 or 2 words about dy/dx, agree?
              – Peter Szilas
              Nov 28 at 9:58










            • Yeah @Peter , should I explain why the approach works or something like that?
              – Akash Roy
              Nov 28 at 10:05










            • Akash.Your approach is fine.But solving for dy/dx you need to divide by 2(1-y) not =0, division by zero is not defined. And we want y=1! Solve for dy/dx, point out that for y=1, dy/dx is not defined, or something like that(my answer).Is this ok?
              – Peter Szilas
              Nov 28 at 10:23


















            • I would kindly ask the down voter what is wrong
              – Akash Roy
              Nov 28 at 8:53










            • Akash.Do not worry about down votes.But: Substitute x=2, y=1......,you can say 1 or 2 words about dy/dx, agree?
              – Peter Szilas
              Nov 28 at 9:58










            • Yeah @Peter , should I explain why the approach works or something like that?
              – Akash Roy
              Nov 28 at 10:05










            • Akash.Your approach is fine.But solving for dy/dx you need to divide by 2(1-y) not =0, division by zero is not defined. And we want y=1! Solve for dy/dx, point out that for y=1, dy/dx is not defined, or something like that(my answer).Is this ok?
              – Peter Szilas
              Nov 28 at 10:23
















            I would kindly ask the down voter what is wrong
            – Akash Roy
            Nov 28 at 8:53




            I would kindly ask the down voter what is wrong
            – Akash Roy
            Nov 28 at 8:53












            Akash.Do not worry about down votes.But: Substitute x=2, y=1......,you can say 1 or 2 words about dy/dx, agree?
            – Peter Szilas
            Nov 28 at 9:58




            Akash.Do not worry about down votes.But: Substitute x=2, y=1......,you can say 1 or 2 words about dy/dx, agree?
            – Peter Szilas
            Nov 28 at 9:58












            Yeah @Peter , should I explain why the approach works or something like that?
            – Akash Roy
            Nov 28 at 10:05




            Yeah @Peter , should I explain why the approach works or something like that?
            – Akash Roy
            Nov 28 at 10:05












            Akash.Your approach is fine.But solving for dy/dx you need to divide by 2(1-y) not =0, division by zero is not defined. And we want y=1! Solve for dy/dx, point out that for y=1, dy/dx is not defined, or something like that(my answer).Is this ok?
            – Peter Szilas
            Nov 28 at 10:23




            Akash.Your approach is fine.But solving for dy/dx you need to divide by 2(1-y) not =0, division by zero is not defined. And we want y=1! Solve for dy/dx, point out that for y=1, dy/dx is not defined, or something like that(my answer).Is this ok?
            – Peter Szilas
            Nov 28 at 10:23


















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