$\text{ If the vertex of the conic represented by }25\left(x^2+y^2\right)=(3x-4y+12{)}^2\text{ is }(a,b\left)\text{ then the value of }\right(b+a)\text{ is }$

$\begin{array}{r}25\left(x^2+y^2\right)=(3x-4y+12{)}^2\\ \Rightarrow \sqrt{x^2+y^2}=\left|\frac{3x-4y+12}{5}\right|\end{array}$

which is equation of the parabola having focus at A(0, 0) and directrix

$L\equiv 3x-4y+12=0-----\left(i\right)$

$\text{ Now equation of axis of the parabola is }y=-\frac{4}{3}x$

$\text{ or }4x+3y=0----\left(ii\right)$

$\text{ Solving (i) and (ii), we get point }B(-36/25,48/25)$

$\text{ Now vertex is the mid point of }AB\text{ , which is }(-18/25,24/25)\text{ . }$