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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?
$begingroup$
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
edited 3 hours ago
Key Flex
8,63761233
asked 3 hours ago
JamesJames
555
$endgroup$
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
edited 3 hours ago
Key Flex
8,63761233
asked 3 hours ago
JamesJames
555
$endgroup$
2
$begingroup$
In short: complete the square
$endgroup$
– Minus One-Twelfth
3 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
edited 3 hours ago
Key Flex
8,63761233
asked 3 hours ago
JamesJames
555
$endgroup$
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
edited 3 hours ago
Key Flex
8,63761233
asked 3 hours ago
JamesJames
555
edited 3 hours ago
Key Flex
8,63761233
asked 3 hours ago
JamesJames
555
edited 3 hours ago
Key Flex
8,63761233
edited 3 hours ago
Key Flex
8,63761233
edited 3 hours ago
Key Flex
8,63761233
8,63761233
asked 3 hours ago
JamesJames
555
asked 3 hours ago
JamesJames
555
asked 3 hours ago
JamesJames
555
555
2
$begingroup$
In short: complete the square
$endgroup$
– Minus One-Twelfth
3 hours ago
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that $dfrac(x-h)^2a^2-dfrac(y-k)^2b^2=1$ is the standard form of hyperbola.
$$9x^2-4y^2-72x=0$$
$$9(x^2-8x)-4y^2=0$$
$$(x^2-8x)-dfrac49y^2=0$$
$$dfrac14(x^2-8x)-dfrac19y^2=0$$
$$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
$$dfrac14(x-4)^2-dfrac19y^2=4$$
$$dfrac(x-4)^216-dfracy^236=1$$
$$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$
answered 3 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
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
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 $$9(x-4)^2-4y^2=144;;;;/:144$$
$$(x-4)^2over 16-y^2over 36=1$$
answered 3 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
3 hours ago
add a comment |
$begingroup$
$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac9144(x-4)^2-frac4144y^2=1$$
$$iff frac(x-4)^216-fracy^236=1$$
$$iff frac(x-4)^24^2-fracy^26^2=1$$
answered 3 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $dfrac(x-h)^2a^2-dfrac(y-k)^2b^2=1$ is the standard form of hyperbola.
$$9x^2-4y^2-72x=0$$
$$9(x^2-8x)-4y^2=0$$
$$(x^2-8x)-dfrac49y^2=0$$
$$dfrac14(x^2-8x)-dfrac19y^2=0$$
$$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
$$dfrac14(x-4)^2-dfrac19y^2=4$$
$$dfrac(x-4)^216-dfracy^236=1$$
$$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$
answered 3 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
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
add a comment |
$begingroup$
Note that $dfrac(x-h)^2a^2-dfrac(y-k)^2b^2=1$ is the standard form of hyperbola.
$$9x^2-4y^2-72x=0$$
$$9(x^2-8x)-4y^2=0$$
$$(x^2-8x)-dfrac49y^2=0$$
$$dfrac14(x^2-8x)-dfrac19y^2=0$$
$$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
$$dfrac14(x-4)^2-dfrac19y^2=4$$
$$dfrac(x-4)^216-dfracy^236=1$$
$$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$
answered 3 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
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
add a comment |
$begingroup$
Note that $dfrac(x-h)^2a^2-dfrac(y-k)^2b^2=1$ is the standard form of hyperbola.
$$9x^2-4y^2-72x=0$$
$$9(x^2-8x)-4y^2=0$$
$$(x^2-8x)-dfrac49y^2=0$$
$$dfrac14(x^2-8x)-dfrac19y^2=0$$
$$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
$$dfrac14(x-4)^2-dfrac19y^2=4$$
$$dfrac(x-4)^216-dfracy^236=1$$
$$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$
answered 3 hours ago
Key FlexKey Flex
8,63761233
$endgroup$
Note that $dfrac(x-h)^2a^2-dfrac(y-k)^2b^2=1$ is the standard form of hyperbola.
$$9x^2-4y^2-72x=0$$
$$9(x^2-8x)-4y^2=0$$
$$(x^2-8x)-dfrac49y^2=0$$
$$dfrac14(x^2-8x)-dfrac19y^2=0$$
$$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
$$dfrac14(x-4)^2-dfrac19y^2=4$$
$$dfrac(x-4)^216-dfracy^236=1$$
$$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$
answered 3 hours ago
Key FlexKey Flex
8,63761233
answered 3 hours ago
Key FlexKey Flex
8,63761233
answered 3 hours ago
Key FlexKey Flex
8,63761233
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
add a comment |
$begingroup$
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
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 $$9(x-4)^2-4y^2=144;;;;/:144$$
$$(x-4)^2over 16-y^2over 36=1$$
answered 3 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
3 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 $$9(x-4)^2-4y^2=144;;;;/:144$$
$$(x-4)^2over 16-y^2over 36=1$$
answered 3 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
3 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 $$9(x-4)^2-4y^2=144;;;;/:144$$
$$(x-4)^2over 16-y^2over 36=1$$
answered 3 hours ago
Maria MazurMaria Mazur
48k1260120
$endgroup$
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 $$9(x-4)^2-4y^2=144;;;;/:144$$
$$(x-4)^2over 16-y^2over 36=1$$
answered 3 hours ago
Maria MazurMaria Mazur
48k1260120
answered 3 hours ago
Maria MazurMaria Mazur
48k1260120
answered 3 hours ago
Maria MazurMaria Mazur
48k1260120
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
add a comment |
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
add a comment |
$begingroup$
$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac9144(x-4)^2-frac4144y^2=1$$
$$iff frac(x-4)^216-fracy^236=1$$
$$iff frac(x-4)^24^2-fracy^26^2=1$$
answered 3 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
$endgroup$
add a comment |
$begingroup$
$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac9144(x-4)^2-frac4144y^2=1$$
$$iff frac(x-4)^216-fracy^236=1$$
$$iff frac(x-4)^24^2-fracy^26^2=1$$
answered 3 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
$endgroup$
add a comment |
$begingroup$
$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac9144(x-4)^2-frac4144y^2=1$$
$$iff frac(x-4)^216-fracy^236=1$$
$$iff frac(x-4)^24^2-fracy^26^2=1$$
answered 3 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
$endgroup$
$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac9144(x-4)^2-frac4144y^2=1$$
$$iff frac(x-4)^216-fracy^236=1$$
$$iff frac(x-4)^24^2-fracy^26^2=1$$
answered 3 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
answered 3 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
answered 3 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
answered 3 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
1,20729
add a comment |
add a comment |
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2
$begingroup$
In short: complete the square
$endgroup$
– Minus One-Twelfth
3 hours ago