Question 129373
{{{3y^2 + 2x - 5y - 4 = 0}}} Start with the given equation


{{{3y^2-5y-4 = -2x}}} Subtract 2x from both sides


{{{3y^2-5y  =-2x+ 4}}} Add 4 to both sides


{{{3(y^2-(5/3)y)  = - 2x+ 4}}} Factor out the leading coefficient 3



Take half of {{{-5/3}}} to get {{{-5/6}}}. Now square {{{-5/6}}} to get {{{25/36}}}


{{{3(y^2-(5/3)y+25/36-25/36)  = - 2x+ 4}}}  Complete the square by adding and subtracting {{{25/36}}}. Note:  {{{25/36-25/36=0}}}


{{{3((y-5/6)^2-25/36)  = - 2x+ 4}}}  Factor {{{y^2-(5/3)y+25/36}}} to get {{{(y-5/6)^2}}}


{{{3(y-5/6)^2-25/12  = - 2x+ 4}}}  Distribute and multiply


{{{3(y-5/6)^2-25/12-4  = - 2x}}}  Subtract 4 from both sides 


{{{3(y-5/6)^2-73/12= - 2x}}}  Combine like terms


{{{(3(y-5/6)^2-73/12)/(-2)=x}}}  Divide both sides by -2 to isolate x


{{{-(3/2)(y-5/6)^2+73/24=x}}}  Divide and reduce


{{{x=-(3/2)(y-5/6)^2+73/24}}}  Rearrange the equation



So after completing the square, the equation goes from {{{3y^2 + 2x - 5y - 4 = 0}}}  to {{{x=-(3/2)(y-5/6)^2+73/24}}}



So the conic section is a parabola that looks like this 



{{{ graph( 500, 500, -10, 10, -10, 10, ((x-73/24)/(-3/2))^(1/2)+5/6,-((x-73/24)/(-3/2))^(1/2)+5/6) }}} Graph of  {{{x=-(3/2)(y-5/6)^2+73/24}}}