Question 968527
{{{4y^2+6x-2y+3x^2-15=2y^2+4x^2}}}
Does this solution process contain a mistake?
{{{3x^2+6x+4y^2-2y-15-2y^2-4x^2=0}}}
{{{-x^2+6x+2y^2-2y-15=0}}}
{{{x^2-6x-2y^2+2y+15=0}}}
Complete The Squares for x and y
{{{x^2-6x+9-2(y^2-y)+15=9}}}
{{{(x-3)^2-2(y^2-y+1/4)+15=9+(-1/2)}}}
{{{(x-3)^2-2(y-1/2)^2+15=8&1/2}}}
{{{(x-3)^2-2(y-1/2)^2=8.5-15=-6.5}}}
{{{-(x-3)^2+2(y-1/2)^2=6.5}}}
{{{-(x-3)^2/6.5+(y-1/2)^2/(6.5/2)=1}}}
Possible mistake but at least steps still lead to hyperbola with vertex on the y axis but not on the x-axis.


Sign mistake, MAYBE made.
A way to avoid that, if it actually happened, could be to start with all x on one side and all y on the other side, and then perform Completion of the Squares.


{{{4y^2+6x-2y+3x^2-15=2y^2+4x^2}}}
{{{4y^2-2y-15=-6x-3x^2+4x^2}}}
{{{4y^2-2y-15=x^2-6x}}}
{{{4(y^2-(1/2)y)-15=x^2-6}}}
{{{4(y^2-(1/2)y+1/4)-15-4(1/4)=(x^2-6x+9-9)}}}
{{{4(y-1/2)^2-15-1=(x-3)^2-9}}}
{{{4(y-1/2)^2-16=(x-3)^2-9}}}
{{{4(y-1/2)^2-(x-3)^2=-9+16}}}
{{{4(y-1/2)^2-(x-3)^2=7}}}
{{{(4/7)(y-1/2)^2-(x-7)^2/7=1}}}
{{{(y-1/2)^2/(7/4)-(x-7)^2/7=1}}}-----Without checking through a solution a third time, this one is probably correct for less likely to have made arithmetic mistakes.
Still hyperbola with vertices on the y axis IF were in a form centered at the origin; but this one is centered at ( 7, 1/2 ).