The coordinate axes are rotated through an angle $\theta$ about the origin in anticlockwise sense. If the equation $2x^2+3xy–6x+2y–4=0$change to $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ then a+b is equal to

Given original equation

 $2{\mathrm{x}}^2+3\mathrm{xy}-6\mathrm{x}+2\mathrm{y}-4=0$

Now transformed equation 

$$\begin{gathered} 2{(X\mathrm{cos\theta }-Y \mathrm{sin\theta })}^2+3(X \mathrm{cos\theta }-Y\mathrm{sin\theta })(X \mathrm{sin\theta }+Y\mathrm{cos\theta }) \\ -6(X \mathrm{cos\theta }-Y\mathrm{sin\theta })+2(X \sin \mathrm{\theta }+Y \mathrm{cos\theta })-4=0 \\ \mathrm{compare} \mathrm{with} {\mathrm{ax}}^2+2\mathrm{h} \mathrm{xy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \\ \mathrm{Now} \mathrm{a}+\mathrm{b}=X^2 \mathrm{coeff}+Y^2 \mathrm{coeff} \\ =(2 {\cos }^2\mathrm{\theta }+3 \mathrm{sin\theta } \mathrm{cos\theta })+(2 {\sin }^2\mathrm{\theta }-3 \sin \mathrm{\theta } \mathrm{cos\theta }) \\ =2({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }) \\ =2\left(1\right) \\ =2 \end{gathered}$$