1. We are given a parabola with equation $y = x^2 + 2x$ that passes through the origin $(0,0)$. 2. We consider a line through the origin with slope $\frac{9}{4}$, so its equation is $y = \frac{9}{4}x$. 3. To find the intersection points of the line and the parabola, set their $y$ values equal: $$x^2 + 2x = \frac{9}{4}x$$ 4. Rearrange the equation: $$x^2 + 2x - \frac{9}{4}x = 0$$ 5. Combine like terms: $$x^2 + \left(2 - \frac{9}{4}\right)x = 0$$ 6. Simplify the coefficient: $$2 - \frac{9}{4} = \frac{8}{4} - \frac{9}{4} = -\frac{1}{4}$$ So the equation becomes: $$x^2 - \frac{1}{4}x = 0$$ 7. Factor out $x$: $$x\left(x - \frac{1}{4}\right) = 0$$ 8. The solutions are: $$x = 0 \quad \text{or} \quad x = \frac{1}{4}$$ 9. The point $x=0$ corresponds to the origin, which we already know. 10. Find the $y$-coordinate of the second intersection point $P$ by substituting $x=\frac{1}{4}$ into the line equation: $$y = \frac{9}{4} \times \frac{1}{4} = \frac{9}{16}$$ 11. Therefore, the $y$-coordinate of point $P$ is $\frac{9}{16}$. Final answer: $\boxed{\frac{9}{16}}$