At what point the origin be shifted so that the equationx2+xy−3x−y+2=0does not contain any first degree term and constant term ?
At what point the origin be shifted so that the equationx2+xy−3x−y+2=0does not contain any first degree term and constant term ?
We have,x2+xy−3x−y+2=0 ...(i)Let the origin be shifted to (h,k).Then x=X+h and y=Y+ksubstituting x=X+h,y=Y+kIn the equation (i),we get(X+h)2+(X+h)(Y+k)−3(X+h)−(Y+k)+2=0⇒X2+h2+2Xh+XY+Xk+Yh+hk−3X−3h−Y−k+2=0⇒X2+XY+2Xh+XK+Yh−2X−Y+h2+hk−3h−k+2=0⇒X2+(2h+k−3)X+XY+9h−1Y+(h2+hk−3h−k+2)=0For this equation to be free from first degree and the constant term,we must have,2h+k−3=0 ...(ii)h−1=0⇒h+1 ...(iii)andh2+hk−3h−h+2=0 ...(iv)Putting h=1 in equation (ii),we get 2+k−3=0⇒k=1Putting h=1 and k=1 in equation (iv),we get(1)2+1−3−1+2=0Hence,the value of h and k satisties the equation(iv)∴The origin is shifted at the point (1,1)
We have,x2+xy−3x−y+2=0 ...(i)Let the origin be shifted to (h,k).Then x=X+h and y=Y+ksubstituting x=X+h,y=Y+kIn the equation (i),we get(X+h)2+(X+h)(Y+k)−3(X+h)−(Y+k)+2=0⇒X2+h2+2Xh+XY+Xk+Yh+hk−3X−3h−Y−k+2=0⇒X2+XY+2Xh+XK+Yh−2X−Y+h2+hk−3h−k+2=0⇒X2+(2h+k−3)X+XY+9h−1Y+(h2+hk−3h−k+2)=0For this equation to be free from first degree and the constant term,we must have,2h+k−3=0 ...(ii)h−1=0⇒h+1 ...(iii)andh2+hk−3h−h+2=0 ...(iv)Putting h=1 in equation (ii),we get 2+k−3=0⇒k=1Putting h=1 and k=1 in equation (iv),we get(1)2+1−3−1+2=0Hence,the value of h and k satisties the equation(iv)∴The origin is shifted at the point (1,1)
