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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
    3 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
    3 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 3 hours ago









Key Flex

8,63761233




8,63761233










asked 3 hours ago









JamesJames

555




555







  • 2




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













  • 2




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








2




2




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





$begingroup$
In short: complete the square
$endgroup$
– Minus One-Twelfth
3 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
    3 hours ago










  • $begingroup$
    @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
    $endgroup$
    – Key Flex
    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
    3 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
      3 hours ago










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










    • $begingroup$
      @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
      $endgroup$
      – Key Flex
      2 hours 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 3 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
      3 hours ago










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










    • $begingroup$
      @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
      $endgroup$
      – Key Flex
      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
    3 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
    3 hours ago












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




    $begingroup$
    @James $dfrac(x-4)^24^2-dfrac(y-0)^26^2$ is in the standard form.
    $endgroup$
    – Key Flex
    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
      3 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
      3 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 3 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
      3 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
      3 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
    3 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
    3 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 3 hours ago









        HAMIDINE SOUMAREHAMIDINE SOUMARE

        1,20729




        1,20729



























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            Torre de la Isleta Índice Véase también Referencias Bibliografía Enlaces externos Menú de navegación38°25′58″N 0°23′02″O / 38.43277778, -0.3838888938°25′58″N 0°23′02″O / 38.43277778, -0.38388889Torre de la Illeta de l’Horta o Torre Saleta. Base de datos de bienes inmuebles. Patrimonio Cultural. Secretaría de Estado de CulturaFicha BIC Torre de la Illeta de l’Horta. Dirección General de Patrimonio Cultural. Generalitat ValencianaLugares de interés. Ayuntamiento del CampelloTorre de la Isleta en CastillosNet.org

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            Bienes de interés cultural de la provincia de AlicanteTorres defensivas de la provincia de AlicanteCampo de AlicanteTorres defensivas de la Comunidad Valenciana en el siglo XVICampello siglo XVICampelloprovincia de AlicanteEspañayacimiento arqueológico de La IsletabóvedamamposteríasillarejosilleríaménsulasVirrey de ValenciaBernardino de Cárdenas y Pachecoduque de Maquedapiratas berberiscosCastillo de Santa BárbaraAlicanteTorre de AguasCampelloBien de Interés Cultural Torre de la Isleta Torre de la Illeta Bien de Interés Cultural Patrimonio histórico de España Vista de la Torre Declaración 12 de febrero de 1996 Figura de protección Monumento Código RI-51 - 0009165 - 00000 [ 1 ] ​ Coordenadas 38°25′58″N 0°23′02″O  /  38.43277778, -0.38388889 Coordenadas: 38°25′58″N 0°23′02″O  /  38.43277778, -0.38388889 ...
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