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How to rewrite equation of hyperbola in standard form


Rewrite a west to east parabola in standard formStandard form of hyperbolaConic Section IntuitionWhat steps are involved to derive a functional expression for the revolving line of a cooling tower?Conic section General form to Standard form HyperbolaHyperbola Standard Form Denominator RelationshipHyperbola with Perpendicular AsymptotesRewrite hyperbola $Ax^2+Bxy+Dx+Ey+F=0$ into standard formHow to prove that the limit of this sequence is $400/pi$Can you multiply an integral by f(x)/f(x) where deg(f(x))>0?













2












$begingroup$


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!






calculus conic-sections







share|cite|improve this question











$endgroup$







  • 2




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
















2












$begingroup$


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!






calculus conic-sections







share|cite|improve this question











$endgroup$







  • 2




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














2












2








2





$begingroup$


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!






calculus 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!






calculus conic-sections




calculus conic-sections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Key Flex

8,63761233




8,63761233










asked 2 hours ago









JamesJames

555




555







  • 2




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













  • 2




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








2




2




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





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











3 Answers
3






active

oldest

votes


















4












$begingroup$

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 hours ago










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









$endgroup$












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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 hours ago










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















    4












    $begingroup$

    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 hours ago










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













    4












    4








    4





    $begingroup$

    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 hours ago









    Key FlexKey Flex

    8,63761233




    8,63761233











    • $begingroup$
      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 hours ago










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










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












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




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









    Maria MazurMaria Mazur

    48k1260120




    48k1260120







    • 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 hours 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 hours 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 hours 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 hours 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 hours ago









        HAMIDINE SOUMAREHAMIDINE SOUMARE

        1,20929




        1,20929



























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