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.
 A: Note that the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ with respect to the line $x-y-2=0$, which has a tangent angle of 45-degrees, is the ellipse that is centered at $(5,2)$, the reflection point of $(4,3)$, and is rotated at 90-degrees, as shown in the graph

So, the equation of the reflection ellipse is,
$$\frac{(x-5)^2}{9}+\frac{(y-2)^2}{16}=1$$
Then, compare with the given equation $16x^2+9y^2+k_1x-36y+k_2=0$ to obtain $k_1=-160$ and $k_2=292$.
A: 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?
A: $$\frac{(x-4)^2}{16}+\frac{(y-3)}{9}=1 \implies x=4 \cos t+4,y=3 \sin t+3 ~~~(1).$$ The image $(X,Y)$ of $(x,y)$ in the line $ax+by+c=0$ is given by:
$$\frac{X-x}{a}=\frac{Y-y}{b}=-2\frac{(ax+by+c)}{a^2+b^2}$$
So we get $$\frac{X-4\cos t-4}{1}=\frac{Y-3\sin t-3}{-1}=-2\frac{4\cos t+4-3\sin t -3-2}{2}$$, we get
$$X=-4\cos t-4 +4\cos t+4+3\sin t+3+2=3 \sin t+5$$ $$ Y=4 \cos t+4-2=  4\cos t +2$$
$$\implies \sin t=\frac{X-5}{3}, ~~ \cos t=\frac{Y-2}{4}$$
Squaring and adding these two results we get the required image the ellipse (1) as
$$\frac{(X-5)^2}{9}+\frac{(Y-2)^2}{16}=1.$$
Finally, one would write the image ellipse as $$\frac{(x-5)^2}{9}+\frac{(y-2)^2}{16}=1.$$ in the same plane as (1)
A: WLOG any point on the ellipse be $P(4+4\cos t, 3+3\sin t )$
If the reflected point is $Q(h,k)$
The midpoint of $PQ$ will lie on $$x-y-2=0$$
As $PQ\perp$ to $x-y-2=0,$ the gradient of $PQ$ will be $$\dfrac{-1}1$$
So, we have two simultaneous equations in $h,k$
Solve them and eliminate $t$ using $$\cos^2t+\sin^2t=1$$
