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Mathematics 7 Online
OpenStudy (anonymous):

put this equation 5x^2-4y^2-40x-16y=36 into the form of y^2/b^2-x^2/a^2=1 or x^2/a^2-y^2/b^2=1

OpenStudy (anonymous):

note: i can't go on neither vyew or skype because of technical difficulites

OpenStudy (anonymous):

\[ 5x^2-4y^2-40x-16y=36 \]

OpenStudy (anonymous):

yes@nodata

OpenStudy (anonymous):

put all the math inside this \[ and

OpenStudy (anonymous):

5x^2-4y^2-40x-16y=36 5(x^2-8x+16)-4(y^2+16y+64)=36-64+(16*5)

OpenStudy (anonymous):

\] this to make it look good

OpenStudy (anonymous):

i dont see the reason to make the equation "look good"

OpenStudy (anonymous):

its already ugly

OpenStudy (anonymous):

5(x-4)^2-4(y+8)^2=36-64+(16*5)

OpenStudy (anonymous):

Because we can understand it better.

OpenStudy (anonymous):

ummm..idk@nodata oh..so u completed the square?@big imran

OpenStudy (anonymous):

yes, that's what I did

OpenStudy (anonymous):

5(x-4)^2-4(y+8)^2=52 divide across by 52

OpenStudy (anonymous):

\[\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2}=1 \]

OpenStudy (anonymous):

btw, I did something wrong arithmetically

OpenStudy (anonymous):

wat?

OpenStudy (anonymous):

I think you can put it in that form if you complete the square

OpenStudy (anonymous):

lol..im kinda confused

OpenStudy (anonymous):

Sorry, It's a good a idea to complete the square.

OpenStudy (anonymous):

And imranmeah91 is working on that.

OpenStudy (anonymous):

5x^2-4y^2-40x-16y=36 5(x^2-8x+16)-4(y^2+4y+4)=100 5(x-4)^2 -4(y+2)^2=100

OpenStudy (anonymous):

wait..where did the 100 com outta from?

OpenStudy (anonymous):

36+80-16=100

OpenStudy (anonymous):

i see where the 36 and 16 came outta..jus the 80

OpenStudy (anonymous):

\[\underline{5}(x^2-8x+\underline{16})-4(y^2+4y+4)=100\]

OpenStudy (anonymous):

that's where 80 come from

OpenStudy (anonymous):

oh..lols im a slow learner..sry

OpenStudy (anonymous):

\[\frac{5(x-4)^2}{100}-4\frac{(y+2)^2}{100}=1\]

OpenStudy (anonymous):

is tht it? can we simplify anymore?

OpenStudy (anonymous):

\[\frac{(x-4)^2}{20}-\frac{(y+2)^2}{25}=1\]

OpenStudy (anonymous):

kay...u simplify?

OpenStudy (anonymous):

since they want it in a^2 , b^2 form \[\frac{(x-4)^2}{\sqrt{20}^2}-\frac{(y+2)^2}{5^2}=1\]

OpenStudy (anonymous):

and the x-4 and y+2 stay, or can they come outta the equation as well?

OpenStudy (anonymous):

no , they stay put

OpenStudy (anonymous):

okay! thank u soo much!!!!

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