If the equation of a curve on reflection of $\frac{{(x-4)}^2}{16}+\frac{{(y-3)}^2}{9}=1$ about the line $x-y=2$ is $16x^2+9y^2+k_{1}x-36y+k_2=0$, then $(k_1,k_2)$ is

Let $P(4+4\cos \theta ,3+3\sin \theta )$ be any point on the ellipse. Then image of P w.r.t line $x-y=2$ is  $\frac{x-4-4\cos \theta }{1}=\frac{y-3-3\sin \theta }{-1}=\frac{-2[4(1+\cos \theta )-3(1+\sin \theta )-2]}{2}$

$\Rightarrow x=5+3\sin \theta \&y=2+4\cos \theta \Rightarrow {\left(\frac{x-5}{3}\right)}^2+{\left(\frac{y-2}{4}\right)}^2=0$

$K_1=-160\&K_2=292$