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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

edited 2 hours ago

Key Flex

8,63761233

asked 2 hours ago

JamesJames

555

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 2 hours ago

Key Flex

8,63761233

asked 2 hours ago

JamesJames

555

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

edited 2 hours ago

Key Flex

8,63761233

asked 2 hours ago

JamesJames

555

$endgroup$

share|cite|improve this question

edited 2 hours ago

Key Flex

8,63761233

asked 2 hours ago

JamesJames

555

share|cite|improve this question

edited 2 hours ago

Key Flex

8,63761233

asked 2 hours ago

JamesJames

555

edited 2 hours ago

Key Flex

8,63761233

edited 2 hours ago

Key Flex

8,63761233

asked 2 hours ago

JamesJames

555

asked 2 hours ago

JamesJames

555

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

answered 2 hours ago

Key FlexKey Flex

8,63761233

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 2 hours ago

Maria MazurMaria Mazur

48k1260120

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

answered 2 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,20929

$endgroup$

add a comment | 

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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

answered 2 hours ago

Key FlexKey Flex

8,63761233

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 2 hours ago

Key FlexKey Flex

8,63761233

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 2 hours ago

Key FlexKey Flex

8,63761233

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Key FlexKey Flex

8,63761233

answered 2 hours ago

Key FlexKey Flex

8,63761233

answered 2 hours ago

Key FlexKey Flex

8,63761233

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

answered 2 hours ago

Maria MazurMaria Mazur

48k1260120

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

answered 2 hours ago

Maria MazurMaria Mazur

48k1260120

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 2 hours ago

Maria MazurMaria Mazur

48k1260120

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Maria MazurMaria Mazur

48k1260120

answered 2 hours ago

Maria MazurMaria Mazur

48k1260120

answered 2 hours ago

Maria MazurMaria Mazur

48k1260120

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

answered 2 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,20929

$endgroup$

add a comment | 

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

answered 2 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,20929

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered 2 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,20929

$endgroup$

share|cite|improve this answer

answered 2 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,20929

answered 2 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,20929

answered 2 hours ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,20929