How to rewrite equation of hyperbola in standard formRewrite 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?
Add big quotation marks inside my colorbox
Can a Canadian Travel to the USA twice, less than 180 days each time?
How to cover method return statement in Apex Class?
What is going on with 'gets(stdin)' on the site coderbyte?
Electoral considerations aside, what are potential benefits, for the US, of policy changes proposed by the tweet recognizing Golan annexation?
Is there a way to get `mathscr' with lower case letters in pdfLaTeX?
Plot of a tornado-shaped surface
How do I delete all blank lines in a buffer?
Quoting Keynes in a lecture
Creepy dinosaur pc game identification
Does Doodling or Improvising on the Piano Have Any Benefits?
What is the highest possible scrabble score for placing a single tile
What are the advantages of simplicial model categories over non-simplicial ones?
Has any country ever had 2 former presidents in jail simultaneously?
Multiplicative persistence
Extract more than nine arguments that occur periodically in a sentence to use in macros in order to typset
Why is the "ls" command showing permissions of files in a FAT32 partition?
Limits and Infinite Integration by Parts
What exact color does ozone gas have?
Does an advisor owe his/her student anything? Will an advisor keep a PhD student only out of pity?
What happens if you are holding an Iron Flask with a demon inside and walk into an Antimagic Field?
How to explain what's wrong with this application of the chain rule?
creating a ":KeepCursor" command
What if a revenant (monster) gains fire resistance?
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 2 hours ago
Key Flex
8,63761233
asked 2 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 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
edited 2 hours ago
Key Flex
8,63761233
asked 2 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 2 hours ago
Key Flex
8,63761233
asked 2 hours ago
JamesJames
555
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
edited 2 hours ago
Key Flex
8,63761233
8,63761233
asked 2 hours ago
JamesJames
555
asked 2 hours ago
JamesJames
555
asked 2 hours ago
JamesJames
555
555
2
$begingroup$
In short: complete the square
$endgroup$
– Minus One-Twelfth
2 hours ago
add a comment |
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
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 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 |
$begingroup$
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 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 |
$begingroup$
$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac9144(x-4)^2-frac4144y^2=1$$
$$iff frac(x-4)^216-fracy^236=1$$
$$iff frac(x-4)^24^2-fracy^26^2=1$$
answered 2 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20929
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158757%2fhow-to-rewrite-equation-of-hyperbola-in-standard-form%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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 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.
$$9x^2-4y^2-72x=0$$
$$9(x^2-8x)-4y^2=0$$
$$(x^2-8x)-dfrac49y^2=0$$
$$dfrac14(x^2-8x)-dfrac19y^2=0$$
$$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
$$dfrac14(x-4)^2-dfrac19y^2=4$$
$$dfrac(x-4)^216-dfracy^236=1$$
$$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$
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.
$$9x^2-4y^2-72x=0$$
$$9(x^2-8x)-4y^2=0$$
$$(x^2-8x)-dfrac49y^2=0$$
$$dfrac14(x^2-8x)-dfrac19y^2=0$$
$$dfrac14(x^2-8x+16)-dfrac19y^2=dfrac14(16)$$
$$dfrac14(x-4)^2-dfrac19y^2=4$$
$$dfrac(x-4)^216-dfracy^236=1$$
$$dfrac(x-4)^24^2-dfrac(y-0)^26^2=1mbox is the required Hyperbola$$
answered 2 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 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
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
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
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
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 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 $$9(x-4)^2-4y^2=144;;;;/:144$$
$$(x-4)^2over 16-y^2over 36=1$$
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 $$9(x-4)^2-4y^2=144;;;;/:144$$
$$(x-4)^2over 16-y^2over 36=1$$
answered 2 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 2 hours ago
Maria MazurMaria Mazur
48k1260120
answered 2 hours ago
Maria MazurMaria Mazur
48k1260120
answered 2 hours ago
Maria MazurMaria Mazur
48k1260120
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
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
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
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 2 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20929
$endgroup$
add a comment |
$begingroup$
$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac9144(x-4)^2-frac4144y^2=1$$
$$iff frac(x-4)^216-fracy^236=1$$
$$iff frac(x-4)^24^2-fracy^26^2=1$$
answered 2 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20929
$endgroup$
add a comment |
$begingroup$
$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac9144(x-4)^2-frac4144y^2=1$$
$$iff frac(x-4)^216-fracy^236=1$$
$$iff frac(x-4)^24^2-fracy^26^2=1$$
answered 2 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20929
$endgroup$
$$9(x^2-8x)-4y^2=9(x-4)^2-144-4y^2=0$$
$$iff frac9144(x-4)^2-frac4144y^2=1$$
$$iff frac(x-4)^216-fracy^236=1$$
$$iff frac(x-4)^24^2-fracy^26^2=1$$
answered 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
answered 2 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20929
1,20929
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158757%2fhow-to-rewrite-equation-of-hyperbola-in-standard-form%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
In short: complete the square
$endgroup$
– Minus One-Twelfth
2 hours ago