If graph of $xy = 1$ is reflected in$y = 2x$ to give new graph whose equation is $12x^2+rxy+s{y}^2+t=0$, then

Let new graph have coordinate (h, k), 

so we have to find locus of (h, k).
Image of (h, k) in y = 2x lies on xy = 1
Let image be $(\alpha ,\beta )$

$\frac{\alpha -h}{2}=\frac{\beta -k}{-1}=\frac{-2(2h-k)}{5}$

$\Rightarrow \alpha =\frac{4k-3h}{5}, \beta =\frac{4h+3k}{5}$

$(\alpha ,\beta )$ lie on xy = 1

$\left(\frac{4k-3h}{5}\right)\left(\frac{4h+3k}{5}\right)=1$

$\Rightarrow 12x^2-12y^2-7xy+25=0$