Whenever we translate axes in a parallel fashion so that the old origin $(0,0)$ becomes $(h,k)$ then any point $(x,y)$ in the old coordinate system becomes $(X,Y)$. So the relation between $(x,y)$ and $(X,Y)$ becomes $$x=X+h$$$$y=Y+k$$
Now lets say we have any curve with the equation $$ax^2+ by^2+kx+ly+c=0$$ with respect to the old axes with origin $(0,0)$ Now, if I change to the new system with origin $(h,k)$ the equation must become $$a(x-h)^2+ b(y-k)^2+k(x-h)+l(y-k)+c=0$$ But my book mentioned it as $$a(x+h)^2+ b(y+k)^2+k(x+h)+l(y+k)+c=0$$ and I am not able to understand which one is correct. Please guide me further.
You can easily perform test.
In old coordinates:
$ f(0,0) = c$
In new coordinates
$ f(-h,-k) = c$
Your book contains right variant. But as for me it uses confusing symbols for variables. Capital letters (X,Y) are more suitable.