Question

If the equation of the curve on reflection of the ellipse $\frac{(x-4{)}^2}{16}+\frac{(y-3{)}^2}{9}=1$ about the line $x-y-2=0$ is $16x^2+9y^2+k_{1}x-36y+k_2=0$, then $\frac{k_1+k_2}{22}$ is

Answer 1

Image of $(h,k)$ on ellipse about $x-y-2=0$ is say $(h^{\prime },k^{\prime })$.
$\frac{h^{\prime }-h}{1}=\frac{k^{\prime }-k}{-1}=-2\frac{h-k-2}{1+1}=-h+k+2$
$\Rightarrow h^{\prime }=k+2,k^{\prime }=h-2$
$(h,k)=(k^{\prime }+2,h^{\prime }-2)$
$\because \frac{(h-4{)}^2}{16}+\frac{(k-3{)}^2}{9}=1\Rightarrow \frac{(k^{\prime }-2{)}^2}{16}+\frac{(h^{\prime }-5{)}^2}{9}=1$
$\Rightarrow 16h^{\prime 2}+9k^{\prime 2}-36k^{\prime }-160h^{\prime }+292=0$
$\Rightarrow$ Equation of reflection of ellipse is
$\Rightarrow 16x^2+6y^2-160x-36y+292=0$
$\therefore \frac{k_1+k_2}{22}=\frac{292-160}{22}=6$