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Mathematics 11 Online
OpenStudy (anonymous):

Put the equation in standard form for an ellipse. 9x^2-54x+16y^2+32y=47

myininaya (myininaya):

its an ellipse!

OpenStudy (anonymous):

by the same method i told for hyperbola: factorise both x and y term. u get 9(x-3)^2+16(y+1)^2=144 therefore, (x-3)^2/16+(y+1)^2/9=1.

myininaya (myininaya):

\[9(x^2-6x)+16(y^2+2y)=47\] \[9(x^2-6x+9)+16(y^2+2y+1)=47+9*9+16*1\] \[9(x-3)^2+16(y+1)^2=144\] \[\frac{(x-3)^2}{16}+\frac{(y+1)^2}{9}=1\]

myininaya (myininaya):

oh it actually does say ellipse in your sentence up there. lol i thought i was pointing something out

OpenStudy (anonymous):

haha ; thanks!!! is it the same way as a hyperbola?

myininaya (myininaya):

yes the only different is you will have a minus sign between instead of a plus

OpenStudy (anonymous):

minus where?

OpenStudy (anonymous):

oh i see, hehe

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