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How to rewrite equation of hyperbola $ 9 x ^ 2 -4y^2-72x=0 $ in standard form [on hold]


How to write this conic equation in standard form?Rewrite a west to east parabola in standard formStandard form of hyperbolaHow so I put these in Standard form? Circle, Ellipse or Hyperbola?Conic Section IntuitionConic section General form to Standard form HyperbolaHyperbola Standard Form Denominator RelationshipHyperbola with Perpendicular AsymptotesRewrite hyperbola $Ax^2+Bxy+Dx+Ey+F=0$ into standard formrewrite expression in another form













1












$begingroup$


I was wondering about this question:




$$ 9 x ^ 2 -4y^2-72x=0 $$




What is the step-by-step process of writing such an equation which, in this case, has the graph of a hyperbola in standard form?



Please excuse me for my messy equation. As I am relatively new to Mathematics Stack Exchange, I do not know how to insert superscripts.



Thank you ahead of time!






conic-sections







share|cite|improve this question











$endgroup$



put on hold as off-topic by Eevee Trainer, Cesareo, Alex Provost, Lee David Chung Lin, max_zorn 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Cesareo, Alex Provost, Lee David Chung Lin, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 3




    $begingroup$
    In short: complete the square
    $endgroup$
    – Minus One-Twelfth
    2 days ago
















1












$begingroup$


I was wondering about this question:




$$ 9 x ^ 2 -4y^2-72x=0 $$




What is the step-by-step process of writing such an equation which, in this case, has the graph of a hyperbola in standard form?



Please excuse me for my messy equation. As I am relatively new to Mathematics Stack Exchange, I do not know how to insert superscripts.



Thank you ahead of time!






conic-sections







share|cite|improve this question











$endgroup$



put on hold as off-topic by Eevee Trainer, Cesareo, Alex Provost, Lee David Chung Lin, max_zorn 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Cesareo, Alex Provost, Lee David Chung Lin, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.











  • 3




    $begingroup$
    In short: complete the square
    $endgroup$
    – Minus One-Twelfth
    2 days ago














1












1








1





$begingroup$


I was wondering about this question:




$$ 9 x ^ 2 -4y^2-72x=0 $$




What is the step-by-step process of writing such an equation which, in this case, has the graph of a hyperbola in standard form?



Please excuse me for my messy equation. As I am relatively new to Mathematics Stack Exchange, I do not know how to insert superscripts.



Thank you ahead of time!






conic-sections







share|cite|improve this question











$endgroup$




I was wondering about this question:




$$ 9 x ^ 2 -4y^2-72x=0 $$




What is the step-by-step process of writing such an equation which, in this case, has the graph of a hyperbola in standard form?



Please excuse me for my messy equation. As I am relatively new to Mathematics Stack Exchange, I do not know how to insert superscripts.



Thank you ahead of time!






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 2 days ago









Maria Mazur

48.4k1260121




48.4k1260121










asked 2 days ago









JamesJames

535




535




put on hold as off-topic by Eevee Trainer, Cesareo, Alex Provost, Lee David Chung Lin, max_zorn 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Cesareo, Alex Provost, Lee David Chung Lin, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.







put on hold as off-topic by Eevee Trainer, Cesareo, Alex Provost, Lee David Chung Lin, max_zorn 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Cesareo, Alex Provost, Lee David Chung Lin, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 3




    $begingroup$
    In short: complete the square
    $endgroup$
    – Minus One-Twelfth
    2 days ago













  • 3




    $begingroup$
    In short: complete the square
    $endgroup$
    – Minus One-Twelfth
    2 days ago








3




3




$begingroup$
In short: complete the square
$endgroup$
– Minus One-Twelfth
2 days ago





$begingroup$
In short: complete the square
$endgroup$
– Minus One-Twelfth
2 days ago











3 Answers
3






active

oldest

votes


















3












$begingroup$

Note that $dfrac(x-h)^2a^2-dfrac(y-k)^2b^2=1$ is the standard form of hyperbola.



$$9x^2-4y^2-72x=0$$
$$9(x^2-8x)-4y^2=0$$
$$(x^2-8x)-dfrac49y^2=0$$
$$dfrac14(x^2-8x)-dfrac19y^2=0$$
$$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
$$dfrac14(x-4)^2-dfrac19y^2=4$$
$$dfrac(x-4)^216-dfracy^236=1$$
$$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Is it not the equation before your answer that is in standard form since the 4^2 and 6^2 become 16 and 36. The equation with 16 & 36 as denominators.
    $endgroup$
    – James
    2 days ago










  • $begingroup$
    @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
    $endgroup$
    – Key Flex
    2 days ago


















2












$begingroup$

So we have $$9(x^2-8x)-4y^2=0$$



$$9(x^2-8x+colorred16-16)-4y^2=0$$



$$9(x-4)^2-144-4y^2=0$$



so $$9(x-4)^2-4y^2=144;;;;/:144$$



$$(x-4)^2over 16-y^2over 36=1$$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    I believe the standard form of a hyperbola involves fractions. I believe the variables are placed as follows: ((x-h)/a^2)-((y-k)/b^2). I may have switched h and k.
    $endgroup$
    – James
    2 days ago


















1












$begingroup$

$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac9144(x-4)^2-frac4144y^2=1$$
$$iff frac(x-4)^216-fracy^236=1$$
$$iff frac(x-4)^24^2-fracy^26^2=1$$






share|cite|improve this answer









$endgroup$



















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Note that $dfrac(x-h)^2a^2-dfrac(y-k)^2b^2=1$ is the standard form of hyperbola.



    $$9x^2-4y^2-72x=0$$
    $$9(x^2-8x)-4y^2=0$$
    $$(x^2-8x)-dfrac49y^2=0$$
    $$dfrac14(x^2-8x)-dfrac19y^2=0$$
    $$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
    $$dfrac14(x-4)^2-dfrac19y^2=4$$
    $$dfrac(x-4)^216-dfracy^236=1$$
    $$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Is it not the equation before your answer that is in standard form since the 4^2 and 6^2 become 16 and 36. The equation with 16 & 36 as denominators.
      $endgroup$
      – James
      2 days ago










    • $begingroup$
      @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
      $endgroup$
      – Key Flex
      2 days ago















    3












    $begingroup$

    Note that $dfrac(x-h)^2a^2-dfrac(y-k)^2b^2=1$ is the standard form of hyperbola.



    $$9x^2-4y^2-72x=0$$
    $$9(x^2-8x)-4y^2=0$$
    $$(x^2-8x)-dfrac49y^2=0$$
    $$dfrac14(x^2-8x)-dfrac19y^2=0$$
    $$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
    $$dfrac14(x-4)^2-dfrac19y^2=4$$
    $$dfrac(x-4)^216-dfracy^236=1$$
    $$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Is it not the equation before your answer that is in standard form since the 4^2 and 6^2 become 16 and 36. The equation with 16 & 36 as denominators.
      $endgroup$
      – James
      2 days ago










    • $begingroup$
      @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
      $endgroup$
      – Key Flex
      2 days ago













    3












    3








    3





    $begingroup$

    Note that $dfrac(x-h)^2a^2-dfrac(y-k)^2b^2=1$ is the standard form of hyperbola.



    $$9x^2-4y^2-72x=0$$
    $$9(x^2-8x)-4y^2=0$$
    $$(x^2-8x)-dfrac49y^2=0$$
    $$dfrac14(x^2-8x)-dfrac19y^2=0$$
    $$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
    $$dfrac14(x-4)^2-dfrac19y^2=4$$
    $$dfrac(x-4)^216-dfracy^236=1$$
    $$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$






    share|cite|improve this answer









    $endgroup$



    Note that $dfrac(x-h)^2a^2-dfrac(y-k)^2b^2=1$ is the standard form of hyperbola.



    $$9x^2-4y^2-72x=0$$
    $$9(x^2-8x)-4y^2=0$$
    $$(x^2-8x)-dfrac49y^2=0$$
    $$dfrac14(x^2-8x)-dfrac19y^2=0$$
    $$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
    $$dfrac14(x-4)^2-dfrac19y^2=4$$
    $$dfrac(x-4)^216-dfracy^236=1$$
    $$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 days ago









    Key FlexKey Flex

    8,61561233




    8,61561233











    • $begingroup$
      Is it not the equation before your answer that is in standard form since the 4^2 and 6^2 become 16 and 36. The equation with 16 & 36 as denominators.
      $endgroup$
      – James
      2 days ago










    • $begingroup$
      @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
      $endgroup$
      – Key Flex
      2 days ago
















    • $begingroup$
      Is it not the equation before your answer that is in standard form since the 4^2 and 6^2 become 16 and 36. The equation with 16 & 36 as denominators.
      $endgroup$
      – James
      2 days ago










    • $begingroup$
      @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
      $endgroup$
      – Key Flex
      2 days ago















    $begingroup$
    Is it not the equation before your answer that is in standard form since the 4^2 and 6^2 become 16 and 36. The equation with 16 & 36 as denominators.
    $endgroup$
    – James
    2 days ago




    $begingroup$
    Is it not the equation before your answer that is in standard form since the 4^2 and 6^2 become 16 and 36. The equation with 16 & 36 as denominators.
    $endgroup$
    – James
    2 days ago












    $begingroup$
    @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
    $endgroup$
    – Key Flex
    2 days ago




    $begingroup$
    @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
    $endgroup$
    – Key Flex
    2 days ago











    2












    $begingroup$

    So we have $$9(x^2-8x)-4y^2=0$$



    $$9(x^2-8x+colorred16-16)-4y^2=0$$



    $$9(x-4)^2-144-4y^2=0$$



    so $$9(x-4)^2-4y^2=144;;;;/:144$$



    $$(x-4)^2over 16-y^2over 36=1$$






    share|cite|improve this answer









    $endgroup$








    • 1




      $begingroup$
      I believe the standard form of a hyperbola involves fractions. I believe the variables are placed as follows: ((x-h)/a^2)-((y-k)/b^2). I may have switched h and k.
      $endgroup$
      – James
      2 days ago















    2












    $begingroup$

    So we have $$9(x^2-8x)-4y^2=0$$



    $$9(x^2-8x+colorred16-16)-4y^2=0$$



    $$9(x-4)^2-144-4y^2=0$$



    so $$9(x-4)^2-4y^2=144;;;;/:144$$



    $$(x-4)^2over 16-y^2over 36=1$$






    share|cite|improve this answer









    $endgroup$








    • 1




      $begingroup$
      I believe the standard form of a hyperbola involves fractions. I believe the variables are placed as follows: ((x-h)/a^2)-((y-k)/b^2). I may have switched h and k.
      $endgroup$
      – James
      2 days ago













    2












    2








    2





    $begingroup$

    So we have $$9(x^2-8x)-4y^2=0$$



    $$9(x^2-8x+colorred16-16)-4y^2=0$$



    $$9(x-4)^2-144-4y^2=0$$



    so $$9(x-4)^2-4y^2=144;;;;/:144$$



    $$(x-4)^2over 16-y^2over 36=1$$






    share|cite|improve this answer









    $endgroup$



    So we have $$9(x^2-8x)-4y^2=0$$



    $$9(x^2-8x+colorred16-16)-4y^2=0$$



    $$9(x-4)^2-144-4y^2=0$$



    so $$9(x-4)^2-4y^2=144;;;;/:144$$



    $$(x-4)^2over 16-y^2over 36=1$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 days ago









    Maria MazurMaria Mazur

    48.4k1260121




    48.4k1260121







    • 1




      $begingroup$
      I believe the standard form of a hyperbola involves fractions. I believe the variables are placed as follows: ((x-h)/a^2)-((y-k)/b^2). I may have switched h and k.
      $endgroup$
      – James
      2 days ago












    • 1




      $begingroup$
      I believe the standard form of a hyperbola involves fractions. I believe the variables are placed as follows: ((x-h)/a^2)-((y-k)/b^2). I may have switched h and k.
      $endgroup$
      – James
      2 days ago







    1




    1




    $begingroup$
    I believe the standard form of a hyperbola involves fractions. I believe the variables are placed as follows: ((x-h)/a^2)-((y-k)/b^2). I may have switched h and k.
    $endgroup$
    – James
    2 days ago




    $begingroup$
    I believe the standard form of a hyperbola involves fractions. I believe the variables are placed as follows: ((x-h)/a^2)-((y-k)/b^2). I may have switched h and k.
    $endgroup$
    – James
    2 days ago











    1












    $begingroup$

    $$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
    $$iff frac9144(x-4)^2-frac4144y^2=1$$
    $$iff frac(x-4)^216-fracy^236=1$$
    $$iff frac(x-4)^24^2-fracy^26^2=1$$






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      $$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
      $$iff frac9144(x-4)^2-frac4144y^2=1$$
      $$iff frac(x-4)^216-fracy^236=1$$
      $$iff frac(x-4)^24^2-fracy^26^2=1$$






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        $$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
        $$iff frac9144(x-4)^2-frac4144y^2=1$$
        $$iff frac(x-4)^216-fracy^236=1$$
        $$iff frac(x-4)^24^2-fracy^26^2=1$$






        share|cite|improve this answer









        $endgroup$



        $$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
        $$iff frac9144(x-4)^2-frac4144y^2=1$$
        $$iff frac(x-4)^216-fracy^236=1$$
        $$iff frac(x-4)^24^2-fracy^26^2=1$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        HAMIDINE SOUMAREHAMIDINE SOUMARE

        1,468211




        1,468211













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