If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is ... 
If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is $16x^2+9y^2+k_1x-36y+k_2=0$, then find $k_1$ and $k_2$

Before solving it, I noticed a problem with it. Even if we reflect the curve, the coefficient of $x^2$ is 9, and not 16 according to what the question says.
That’s all I have as doubt. I don’t need the whole answer, just need to know if the question is right or not.

Consider the line  $y=x-2$;
New coordinates : $X=x$; and $Y=y+2$;
Reflect about $Y=X.$ 
The given curve expressed in $X,Y:$
$(X-4)^2/4^2+(Y-5)^2/3^2=1$;
The reflected curve:
$(Y-4)^2/4^2+(X-5)^2/3^2=1$;
Back to $x,y$:
$(y-2)^2/4^2+(x-5)^2/3^2=1$;
$9(y-2)^2+16(x-5)^2=16\cdot 9$;
Can you finish?
