How to rewrite equation of hyperbola in standard form












2












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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







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  • 2




    $begingroup$
    In short: complete the square
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    – 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






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share|cite|improve this question













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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

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4












$begingroup$

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



$$9x^2-4y^2-72x=0$$
$$9(x^2-8x)-4y^2=0$$
$$(x^2-8x)-dfrac49y^2=0$$
$$dfrac14(x^2-8x)-dfrac19y^2=0$$
$$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac{1}{4}(16)$$
$$dfrac14(x-4)^2-dfrac19y^2=4$$
$$dfrac{(x-4)^2}{16}-dfrac{y^2}{36}=1$$
$$dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}=1mbox{ is the required Hyperbola}$$






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 $dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}$ is in the standard form.
    $endgroup$
    – Key Flex
    1 hour ago



















2












$begingroup$

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



$$9(x^2-8x+color{red}{16-16})-4y^2=0$$



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



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












$begingroup$

$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac{9}{144}(x-4)^2-frac{4}{144}y^2=1$$
$$iff frac{(x-4)^2}{16}-frac{y^2}{36}=1$$
$$iff frac{(x-4)^2}{4^2}-frac{y^2}{6^2}=1$$






share|cite|improve this answer









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



    $$9x^2-4y^2-72x=0$$
    $$9(x^2-8x)-4y^2=0$$
    $$(x^2-8x)-dfrac49y^2=0$$
    $$dfrac14(x^2-8x)-dfrac19y^2=0$$
    $$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac{1}{4}(16)$$
    $$dfrac14(x-4)^2-dfrac19y^2=4$$
    $$dfrac{(x-4)^2}{16}-dfrac{y^2}{36}=1$$
    $$dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}=1mbox{ is the required Hyperbola}$$






    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 $dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}$ is in the standard form.
      $endgroup$
      – Key Flex
      1 hour ago
















    4












    $begingroup$

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



    $$9x^2-4y^2-72x=0$$
    $$9(x^2-8x)-4y^2=0$$
    $$(x^2-8x)-dfrac49y^2=0$$
    $$dfrac14(x^2-8x)-dfrac19y^2=0$$
    $$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac{1}{4}(16)$$
    $$dfrac14(x-4)^2-dfrac19y^2=4$$
    $$dfrac{(x-4)^2}{16}-dfrac{y^2}{36}=1$$
    $$dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}=1mbox{ is the required Hyperbola}$$






    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 $dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}$ is in the standard form.
      $endgroup$
      – Key Flex
      1 hour ago














    4












    4








    4





    $begingroup$

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



    $$9x^2-4y^2-72x=0$$
    $$9(x^2-8x)-4y^2=0$$
    $$(x^2-8x)-dfrac49y^2=0$$
    $$dfrac14(x^2-8x)-dfrac19y^2=0$$
    $$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac{1}{4}(16)$$
    $$dfrac14(x-4)^2-dfrac19y^2=4$$
    $$dfrac{(x-4)^2}{16}-dfrac{y^2}{36}=1$$
    $$dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}=1mbox{ is the required Hyperbola}$$






    share|cite|improve this answer









    $endgroup$



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



    $$9x^2-4y^2-72x=0$$
    $$9(x^2-8x)-4y^2=0$$
    $$(x^2-8x)-dfrac49y^2=0$$
    $$dfrac14(x^2-8x)-dfrac19y^2=0$$
    $$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac{1}{4}(16)$$
    $$dfrac14(x-4)^2-dfrac19y^2=4$$
    $$dfrac{(x-4)^2}{16}-dfrac{y^2}{36}=1$$
    $$dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}=1mbox{ is the required Hyperbola}$$







    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 $dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}$ 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 $dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}$ 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 $dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}$ is in the standard form.
    $endgroup$
    – Key Flex
    1 hour ago




    $begingroup$
    @James $dfrac{(x-4)^2}{4^2}-dfrac{(y-0)^2}{6^2}$ is in the standard form.
    $endgroup$
    – Key Flex
    1 hour ago











    2












    $begingroup$

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



    $$9(x^2-8x+color{red}{16-16})-4y^2=0$$



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



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



    $$9(x^2-8x+color{red}{16-16})-4y^2=0$$



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



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



    $$9(x^2-8x+color{red}{16-16})-4y^2=0$$



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



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



    $$9(x^2-8x+color{red}{16-16})-4y^2=0$$



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



    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












    $begingroup$

    $$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
    $$iff frac{9}{144}(x-4)^2-frac{4}{144}y^2=1$$
    $$iff frac{(x-4)^2}{16}-frac{y^2}{36}=1$$
    $$iff frac{(x-4)^2}{4^2}-frac{y^2}{6^2}=1$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
      $$iff frac{9}{144}(x-4)^2-frac{4}{144}y^2=1$$
      $$iff frac{(x-4)^2}{16}-frac{y^2}{36}=1$$
      $$iff frac{(x-4)^2}{4^2}-frac{y^2}{6^2}=1$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
        $$iff frac{9}{144}(x-4)^2-frac{4}{144}y^2=1$$
        $$iff frac{(x-4)^2}{16}-frac{y^2}{36}=1$$
        $$iff frac{(x-4)^2}{4^2}-frac{y^2}{6^2}=1$$






        share|cite|improve this answer









        $endgroup$



        $$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
        $$iff frac{9}{144}(x-4)^2-frac{4}{144}y^2=1$$
        $$iff frac{(x-4)^2}{16}-frac{y^2}{36}=1$$
        $$iff frac{(x-4)^2}{4^2}-frac{y^2}{6^2}=1$$







        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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