ax2+dy2=1 ⇒cx2+d(1−ax)2=1 ⇒b2cx2+d(1+a2x2−2ax)=b2 ⇒(b2c+a2d)x2−2adx+d−b2=0 both the equations have only one solution ⇒ discriminant = 0 ⇒4a2d2−4(b2c+a2d)(d−b2)=0 ⇒a2d2−(b2c+a2d)(d−b2)=0 ⇒a2d2−b2cd+b4c−a2d2+a2b2d=0 ⇒b4c+a2b2d=b2cd ⇒