Question:

Step 1: Reflect the hyperbola $xy = 1$ in the line $y = 2x$. Use the reflection formula for a point $(x, y)$ in the line $y = mx$ (here $m = 2$): $$x' = \frac{x(1-m^2) + 2my}{1+m^2}, \quad y' = \frac{2mx - y(1-m^2)}{1+m^2}$$ With $m = 2$: $$x' = \frac{x(1-4) + 4y}{5} = \frac{-3x + 4y}{5}, \quad y' = \frac{4x - y(1-4)}{5} = \frac{4x + 3y}{5}$$ Step 2: Express $x$ and $y$ in terms of $x'$ and $y'$ (inverse reflection, same formula since reflection is its own inverse): $$x = \frac{-3x' + 4y'}{5}, \quad y = \frac{4x' + 3y'}{5}$$ Step 3: Substitute into $xy = 1$: $$\frac{(-3x' + 4y')(4x' + 3y')}{25} = 1$$ $$(-3x' + 4y')(4x' + 3y') = 25$$ $$-12x'^2 - 9x'y' + 16x'y' + 12y'^2 = 25$$ $$-12x'^2 + 7x'y' + 12y'^2 = 25$$ $$12x'^2 - 7x'y' - 12y'^2 + 25 = 0$$ Step 4: Compare with $12x^2 + rxy + sy^2 + t = 0$: $$r = -7,\ s = -12,\ t = 25$$ Option (c) is correct. Step 5: Check option (d): $r + s = -7 + (-12) = -19$. Option (d) is correct.