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Divide both sides by 144
as Mertsj suggested already \(\bf 9x^2-16y^2=144\implies \cfrac{9x^2}{144}-\cfrac{16y^2}{144}=1 \\ \quad \\ recall \implies \cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1\)
What conic section is it?
Your 9x^2/144-16y^2/144=1 needs to be re-written in the form that jdoe came up with ("recall"). I'd first type this 9x^2/144-16y^2/144=1 in Equation Editor for added clarity:\[\frac{ 9x^2 }{ 144 }-\frac{ 16y^2 }{ 144 }=1\]
@jamielovesmickey : This doesn't look exactly like jdoe's formula. Can you "fix" it?
So draw your box: The center is (0,0). go 4 units right and 4 units left of center. Go 3 units up and 3 units down from the center.
The asymptotes are the diagonals of the box. Draw them. This hyperbola opens to the right and left since x is positive. Draw it using the asymptotes as guidelines.
yw
@mertsj: mind explaining to Jamie where that "over 4 and up 4" came from?
Where does it say "over 4 and up 4"?
"Go four units up and four units right of center."
Is this what you are referring to: "go 4 units right and 4 units left of center. "
Afraid so. Forgot to put brain in gear before putting fingers into action on the keyboard. Sorry!
np. I know I make my share of mistakes.
My questions stemmed from wondering whether Jamie might need an explanation of where that 3 and that 4 came from. Earlier, I'd suggested that the equation I wrote via Equation Editor be simplified; doing so would make the source of that 3 and that 4 more obvious.
I see.
\[\frac{9x^2}{144}=\frac{x^2}{16}\]
sqrt16=4 That's where the 4 comes from.
\[\frac{16y^2}{144}=\frac{y^2}{9}\] sqrt9 = 3 That's where the 3 comes from.
That's why you divide to make the right side 1 because then a and b can be identified.
yw
In \[\frac{ 9x^2 }{ 144 }-\frac{ 16y^2 }{ 144 }=1\], that 144/9 can be reduced to 16=4^2, and that 144/16 can be reduced to 9=3^2.
I would like to get off the 'Net now, but may be available later this evening. C u!
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