16x2 + 25y2 - 96x - 200y + 144 = 0....can someone please help me put this in standard form?
ellipse right?
yes!
i dont think so....
answer or method? method is a pain, answer is easy you are going to have to complete the square to make it look like \[16(x-h)^2+25(y-k)^2=C\] and then divide to get \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]
\[16x^2 + 25y^2 - 96x - 200y + 144 = 0\] \[16(x^2-6x)+25(y^2-4)=-144\] is your start
then to complete the square take half of 6, square it to get 9 and multiply by 16 to get 144 also take half of 4, which is 2, square to get 4, multiply by 25 to get 100 and write \[16(x-3)^2+25(y-2)^2=-144+144+100\]
oh damn i make a mistake!!
\[16(x^2-6x)+25(y^2-4)=-144\] should be \[16(x^2-6x)+25(y^2-8)=-144\]
ok im following so far..this is great
you could just do answer if you want
ok same idea, this time get \[16(x-3)^2+25(y-4)^2=-144+144+400\]
i made a mistake it should have been 8 instead of 4 therefore half of 8 is 4, 4 squares is 16 and 16 times 25 is 400
so you get \[16(x-3)^2+25(y-4)^2=400\]
your last step is to divide both sides by 400 to put a 1 on the right, and you are done
sorry but what would that look like?
what is \(\frac{16}{400}\)?
1/25
so first term is \(\frac{(x-3)^2}{25}\)
ok got it all
THANKS!
do you know how to find the foci of this?
the center is (3,4)
@satellite73
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