Answer
At $(3,0)$, $\kappa = \dfrac{3}{16}$
At $(0,4)$, $\kappa = \dfrac{4}{9}$
Work Step by Step
The curvature $\kappa$ for a plane curve system is: $\kappa = \dfrac{(x' y''-y'x'')}{(x'+y')^{3/2}}$
Now, $x=3 \cos t \implies x' =-3 \sin t $ and $x'' =-3 \cos t $ and $y =4 \sin t \implies y' =4 \cos t $ and $y'' =-4 \sin t $
Now, $\kappa = \dfrac{(12 \sin^2 t+12 \cos^2 t)}{(9 \sin^2 t 16 \cos^2 t)^{3/2}} =\dfrac{12}{(9 \sin^2 t 16 \cos^2 t)^{3/2}}$
At $(3,0)$, $\kappa = \dfrac{(12 \sin^2 t+12 \cos^2 t)}{(9 \sin^2 t + 16 \cos^2 t)^{3/2}} =\dfrac{12}{16 \cos^2 (0)^{3/2}}=\dfrac{3}{16}$
At $(0,4)$, $\kappa = \dfrac{12}{(9 \sin^2 (\pi/2) + 16 \cos^2 (\pi/2))^{3/2}} =\dfrac{12}{9^{3/2}}=\dfrac{4}{9}$
