Question

16x^2 + 25 y ^ 2 - 32x + 50 y + 31 = 0

Answer 1

combine like terms....first and what do you get?

Answer 2

i get up to there but when i completed the square i was confused

Answer 3

\[16(x ^{2}-2x)+25(y ^{2}+2y)+31=0\]
applying completing the square method::
\[16((x ^{2}-2x+1)-1)+25((y ^{2}+2y+1)-1)+31=0\]
\[16(x-1)^{2}+25(y+1)^{2}-16-25+31=0\]
\[16(x-1)^{2}+25(y+1)^{2}=10\]

Answer 4

perfect thats where i got up to

Answer 5

now divide by 10 on both sides

Answer 6

ya/...

Answer 7

thats a problem

Answer 8

\[(8/5)(x-1)^{2}+(5/2)(y+1)^{2}=1\]
or
\[(x-1)^{2}/(5/8)+(y+1)^{2}/(5/2)=1\] which is an ellipse

Answer 9

ok but what's gonna be a ?

Answer 10

a=\[\sqrt{5/8}\]

Answer 11

how ?

Answer 12

idk