Find the angle at which the curves x² = 4ay and y =

Find the angle at which the curves $x^{2} = 4ay$ and $y = \dfrac{8a^{3}}{x^{2}+4a^{2}}$ intersect, where $a \gt 0$.

Step 1: Find the intersection points

From the parabola $y = \dfrac{x^{2}}{4a}$. Substituting into the second curve and cross-multiplying: \[\frac{x^{2}}{4a} = \frac{8a^{3}}{x^{2}+4a^{2}} \;\Longrightarrow\; x^{2}\bigl(x^{2}+4a^{2}\bigr) = 32a^{4} \;\Longrightarrow\; x^{4}+4a^{2}x^{2}-32a^{4} = 0.\]

With $u = x^{2}$: $u^{2}+4a^{2}u-32a^{4} = 0 \;\Longrightarrow\; u = \frac{-4a^{2}\pm 12a^{2}}{2}$.

and only the positive root $u = 4a^{2}$ is admissible. Hence

\[ x = \pm 2a, \qquad y = \frac{4a^{2}}{4a} = a: \qquad P_{1}(2a,\,a), \quad P_{2}(-2a,\,a). \]

Step 2: Slope of the parabola at $P_{1}$

\[ y = \frac{x^{2}}{4a} \;\Longrightarrow\; y' = \frac{x}{2a}, \qquad k_{1} = \frac{2a}{2a} = 1. \]

Step 3: Slope of the second curve at $P_{1}$

Writing $y = 8a^{3}\bigl(x^{2}+4a^{2}\bigr)^{-1}$ and using the chain rule: \[y' = -\frac{16a^{3}x}{\bigl(x^{2}+4a^{2}\bigr)^{2}}, \qquad k_{2} = -\frac{16a^{3}\cdot 2a}{\bigl(8a^{2}\bigr)^{2}} = -\frac{32a^{4}}{64a^{4}} = -\frac12.\]

Step 4: Apply the angle formula

\[ \tan\alpha = \left|\frac{k_{1}-k_{2}}{1+k_{1}k_{2}}\right| = \left|\frac{1+\tfrac12}{1-\tfrac12}\right| = \frac{\tfrac32}{\tfrac12} = 3, \qquad \alpha = \arctan 3 \approx 71.565^{\circ}. \]

Step 5: The second point

At $P_{2}(-2a,a)$ both slopes reverse: $k_{1} = -1$, $k_{2} = \tfrac12$, and

\[ \tan\alpha = \left|\frac{-1-\tfrac12}{1-\tfrac12}\right| = 3, \]

the same angle, by the symmetry of both curves about the $y$-axis.

Step 6: Verification

\[P_{1} \text{ on the parabola: } (2a)^{2} = 4a\cdot a \;\checkmark; \quad \text{on the second: } \frac{8a^{3}}{4a^{2}+4a^{2}} = a \;\checkmark\]

\[\arctan 3 = 71.565^{\circ} = 71^{\circ} + 0.565\times 60' \approx 71^{\circ}34' \;\;\checkmark\]