1. **Problem statement:** Given a parabola with vertex at the origin and equation $$Ay^2 = (3A - 4)x$$ where $$A \in \mathbb{R}$$, find the value of $$A$$ such that the parabola passes through the point $$\left(-\frac{1}{4}, 3\right)$$. 2. **Formula and approach:** We substitute the point coordinates $$x = -\frac{1}{4}$$ and $$y = 3$$ into the equation to find $$A$$. 3. **Substitute the point:** $$A \cdot 3^2 = (3A - 4) \cdot \left(-\frac{1}{4}\right)$$ Simplify: $$9A = -\frac{1}{4}(3A - 4)$$ 4. **Multiply both sides by 4 to clear the denominator:** $$4 \cdot 9A = 4 \cdot \left(-\frac{1}{4}(3A - 4)\right)$$ $$36A = - (3A - 4)$$ 5. **Distribute the negative sign:** $$36A = -3A + 4$$ 6. **Add $$3A$$ to both sides:** $$36A + 3A = 4$$ $$39A = 4$$ 7. **Solve for $$A$$:** $$A = \frac{4}{39}$$ 8. **Final answer:** The value of $$A$$ is $$\boxed{\frac{4}{39}}$$. This means the parabola equation is: $$\frac{4}{39} y^2 = \left(3 \cdot \frac{4}{39} - 4\right) x$$ or simplified further if needed.