Comparing the given equation with `ax^(2)+by^(2)+2hxy+agx+2fy+c=0`, we have `a=9,b=4,c=3,h=lamda,g=2,andf=-1`. If the equation `9x^(2)+4y^(2)+2lamdaxy+4x-2y+3=0` represents a parabola, then its second-degree terms must form the perfect square. Therefore,
`lamda^(2)=36" ""(Using "h^(2)-ab=0)`
`orlamda=pm6`
Also, for these values of `lamda,Delta!=0`.
