Question

CXan someone help oout please?

Answer 1

convert each equation to standard form and determine the locations of the center vertices co vertices and foci 2x^2+3y^2-8x+6y+5=0

Answer 2

\[2x^{2}+3y^{2}-8x+6y+5=0\]first look at just the terms in x and see if they can be written as (x+a)^2 where "a" is some constant\[2x^{2}-8x=2(x^{2}-4x)=2((x-2)^{2}-4)\]the extra -4 is there because (x-2)^2 expands to x^2-4x+4 so we need to get rid of the +4.
now do the same with the terms in y:\[3y^{2}+6y=3(y^{2}+2y)=3((y+1)^{2}-1)\]again we need the extra -1 because (y+1)^2 expands to y^2+2y+1 so we need to get rid of the +1.
next put these back into original equation:
\[2x^{2}+3y^{2}-8x+6y+5=0\]\[2((x-2)^{2}-4)+3((y+1)^{2}-1)+5=0\]\[2(x-2)^{2}-8+3(y+1)^{2}-3+5=0\]\[2(x-2)^{2}+3(y+1)^{2}-6=0\]\[2(x-2)^{2}+3(y+1)^{2}=6\]now divide both sides by lowest common multiple of the denominators on the LHS (i.e. LCM of 2 and 3 which is 6):\[\frac{(x-2)^{2}}{3}+\frac{(y+1)^{2}}{2}=1\]you should be able to work out the rest from here. I need to go and catch some Zzz's now...