x3_drummerchick
  • x3_drummerchick
how do I put this equation into hyperbola form? 81x^2 -144y^2 = 1 Will give medals!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
x3_drummerchick
  • x3_drummerchick
|dw:1449419286872:dw|
x3_drummerchick
  • x3_drummerchick
can i take the recipricol of the whole thing?
mathmale
  • mathmale
This is already in the form of an equation of a hyperbola: the first term is +, the second is -, and the right side is '1. True, the standard form of the equation of a hyperbola centered at the origin is \[\frac{ x^2 }{ a^2 }-\frac{ y^2 }{ b^2 }=1\]

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mathmale
  • mathmale
Can you transform the given equation into this standard form?
mathmale
  • mathmale
No, you can't "take the reciprocal of the whole thing."
x3_drummerchick
  • x3_drummerchick
can i flip everything so it looks like this? :
mathmale
  • mathmale
Your given expression has 2 terms on the left; your job is to re-write them so as to fit "standard form."
x3_drummerchick
  • x3_drummerchick
|dw:1449419502408:dw|
mathmale
  • mathmale
How would you convert \[\frac{ 81x^2 }{ 1}\]
mathmale
  • mathmale
to the form \[\frac{ x^2 }{ a^2 }?\]
x3_drummerchick
  • x3_drummerchick
im not sure :( my professor never showed us a problem like this. we had a sub
mathmale
  • mathmale
|dw:1449419635455:dw|
x3_drummerchick
  • x3_drummerchick
im not sure what else to do
mathmale
  • mathmale
Hint:\[\frac{ 81x^2 }{ 1^2 }=(\frac{ 9}{ 1 })^2\]
x3_drummerchick
  • x3_drummerchick
where did the y squared go?
mathmale
  • mathmale
Rewrite\[\frac{ 81x^2 }{ 1^2 }\]
x3_drummerchick
  • x3_drummerchick
|dw:1449419844971:dw|
mathmale
  • mathmale
in the form\[\frac{ (fraction)^2 }{ 1 }x^2\]
mathmale
  • mathmale
|dw:1449419898542:dw|
x3_drummerchick
  • x3_drummerchick
ohh, so are just pulling out the terms away from the variable?
mathmale
  • mathmale
In other words, re-write (9/1)^2 in the form \[\frac{ 1 }{ fraction }\]
mathmale
  • mathmale
Pulling the coefficients away from the respective variables, yes. Why? Because we need to re-write \[81x^2\]
x3_drummerchick
  • x3_drummerchick
|dw:1449420038971:dw|
mathmale
  • mathmale
in the form\[[1/fraction]\]
x3_drummerchick
  • x3_drummerchick
|dw:1449420094732:dw|
mathmale
  • mathmale
because the standard form of the equation of a hyp. is\[\frac{ x^2 }{ a^2 }-\frac{ y^2 }{b^2 }=1\]
x3_drummerchick
  • x3_drummerchick
so i sorta had the right idea, i just needed to take the root of the coefficients
mathmale
  • mathmale
|dw:1449420159673:dw|
x3_drummerchick
  • x3_drummerchick
so say these were perfect squares, and the problem was like, 14x^2 - 17y^2=1, would the standard form be
x3_drummerchick
  • x3_drummerchick
|dw:1449420263980:dw|
mathmale
  • mathmale
If the first coeff. is 81, that can be rewritten as \[\frac{ 1 }{ \frac{ 1 }{ 81 } }=\frac{ 1 }{( \frac{ 1 }{ 9 })^2 }\]
x3_drummerchick
  • x3_drummerchick
oh snap, okay. because that value is a squared
mathmale
  • mathmale
Right, and there's no squaring of the denominators in your question involving coeff. 14.
x3_drummerchick
  • x3_drummerchick
so standard form of original equation is :
x3_drummerchick
  • x3_drummerchick
|dw:1449420496067:dw| ?
mathmale
  • mathmale
There's still a problem here because neither y our 9 nor your 12 appears squared.
mathmale
  • mathmale
\[81x^2-144y^2=1 \]
mathmale
  • mathmale
is the given form. It can be re-written as \[\frac{ x^2 }{ \frac{ 1 }{ 81 } }-\frac{ y^2 }{ \frac{ 1 }{ 144 } }=1\]
x3_drummerchick
  • x3_drummerchick
oh i see, the a and b have to be squared, so the original value of a is 1/9, and b is 1/112. squaring them gives us 1/81 and 1/144
mathmale
  • mathmale
You could either stop here, or continue, for example, re-writing 1/81 as \[(\frac{ 1 }{ 9 })^2\]
mathmale
  • mathmale
If you do that, then your ": a " would be 1/9 and your a^2 would be 1/81.
x3_drummerchick
  • x3_drummerchick
i think i actually get it.
mathmale
  • mathmale
Once again, y ou must convert the given equation to the standard form of a hyperbola, which is \[\frac{ x^2 }{ a^2 }-\frac{ y^2 }{ b^2 }=1\]
mathmale
  • mathmale
I'm so glad this makes sense to you now. Thanks for your patience.
x3_drummerchick
  • x3_drummerchick
no no, thank you for your patience and actually EXPLAINING everything to me so I dont just get an answer. you should teach math. :)
x3_drummerchick
  • x3_drummerchick
thank you again so much!
mathmale
  • mathmale
It surely feels good to be appreciated. Thanks for the very kind words. Actually, I'm a retired math teacher. Merry Christmas!
x3_drummerchick
  • x3_drummerchick
my professor always has substitutes, and doesnt explain things well even when he is here :( its definitely evident! merry christmas to you too!
mathmale
  • mathmale
so sorry about your professor's frequent absences. I'll be happy to work with you any time we're both on OpenStudy together.
x3_drummerchick
  • x3_drummerchick
welp, Im pursuing calc next semester so im sure ill be needing a hand lol thanks again!
anonymous
  • anonymous
\[81 x^2-144y^2=1\] \[\frac{ x^2 }{ \left( \frac{ 1 }{ 9 } \right) ^2}-\frac{ y^2 }{ \left( \frac{ 1 }{ 12 } \right)^2 }=1\]

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