Given transformed equation is 2x2+3xy−2y2−17x+6y+8=0 Original equationx=X−α,y=Y−β 2(X−α)2+3(X−α)(Y−β) −2(Y−β)2−17(X−α)+6(Y−β)+8=0 ⇒2(X2+α2−2αX)+3(XY−βX−αY+αβ) −2(Y2+β2−2βY)−17X+17α+6Y −6β+8=0 ⇒2X2+3XY−2Y2+X(−4α−3β−17) +Y(−3α+4β+6) +2α2+3αβ−2β2+17α−6β+8=0 Comparing it with aX2+2hXY+bY2+C=0 We get,−4α−3β−17=0.........(i) −3α+4β+6=0.......(ii) Solving Eqs. (i) and (ii), we get β=−3 and α=−2 C=2α2+3αβ−2β2+17α−6β+8 C=0 ∵3α+C=3(−2)+0=−6=2(−3) =2β
