1. **Problem statement:** Given the equation $$Ay^2 = (3A - 4)x$$ where $A \in \mathbb{R}$, find the value of $A$ such that the curve passes through the point $(3, -\frac{1}{4})$. 2. **Substitute the point into the equation:** We plug $x=3$ and $y=-\frac{1}{4}$ into the equation: $$A\left(-\frac{1}{4}\right)^2 = (3A - 4) \times 3$$ 3. **Simplify the left side:** $$A \times \frac{1}{16} = 3(3A - 4)$$ 4. **Rewrite the equation:** $$\frac{A}{16} = 9A - 12$$ 5. **Bring all terms to one side:** $$\frac{A}{16} - 9A + 12 = 0$$ 6. **Find common denominator and combine:** $$\frac{A}{16} - \frac{144A}{16} + 12 = 0$$ $$\frac{A - 144A}{16} + 12 = 0$$ $$\frac{-143A}{16} + 12 = 0$$ 7. **Isolate $A$:** $$\frac{-143A}{16} = -12$$ 8. **Multiply both sides by 16:** $$\cancel{16} \times \frac{-143A}{\cancel{16}} = -12 \times 16$$ $$-143A = -192$$ 9. **Divide both sides by -143:** $$A = \frac{-192}{-143} = \frac{192}{143}$$ **Final answer:** $$A = \frac{192}{143}$$