曲线 $C$ 由参数方程
$x = \frac{t}{1 + t}, \quad y = \frac{t^2}{1 + t},$
定义，其中 $t$ 取除 $-1$ 以外的所有实数值。 求 $\frac{\text{d}y}{\text{d}x}$，用 $t$ 表示你的答案。

1. Find $\frac{\text{d}y}{\text{d}x}$ in terms of $t$ (required from the previous unnumbered part).

$\frac{\text{d}x}{\text{d}t} = \frac{(1)(1 + t) - t(1)}{(1 + t)^2} = \frac{1}{(1 + t)^2}$

$\frac{\text{d}y}{\text{d}t} = \frac{2t(1 + t) - t^2(1)}{(1 + t)^2} = \frac{t(t + 2)}{(1 + t)^2}$

$\frac{\text{d}y}{\text{d}x} = \frac{\frac{\text{d}y}{\text{d}t}}{\frac{\text{d}x}{\text{d}t}} = t(t + 2)$

2. Determine the gradient of the tangent at $t = p$ and substitute into the equation of a line.

At the point $\left(\frac{p}{1 + p}, \frac{p^2}{1 + p}\right)$, the parameter is $t = p$.

The gradient of the tangent is $m = p(p + 2)$.

Using the point-gradient form $y - y_1 = m(x - x_1)$:
$y - \frac{p^2}{1 + p} = p(p + 2)\left(x - \frac{p}{1 + p}\right)$

3. Simplify the linear equation to arrive at the required result.

$y = p(p + 2)x - \frac{p^2(p + 2)}{1 + p} + \frac{p^2}{1 + p}$

$y = p(p + 2)x - \frac{p^3 + 2p^2 - p^2}{1 + p}$

$y = p(p + 2)x - \frac{p^3 + p^2}{1 + p}$

$y = p(p + 2)x - \frac{p^2(p + 1)}{1 + p}$

Since $p \neq -1$, we can cancel $(1 + p)$:
$y = p(p + 2)x - p^2 \quad \text{(Shown)}$

1. Find the values of $p$ for the tangents that pass through $(2, 5)$.

Substitute $(x, y) = (2, 5)$ into the equation of the tangent $y = p(p + 2)x - p^2$:
$5 = p(p + 2)(2) - p^2$
$5 = 2p^2 + 4p - p^2$
$p^2 + 4p - 5 = 0$
$(p + 5)(p - 1) = 0$
$p = 1 \text{ or } p = -5$

2. Determine the gradients of the two tangents.

From the tangent equation $y = p(p + 2)x - p^2$, the gradient $m$ is $p(p + 2)$.

When $p = 1$,
$m_1 = 1(1 + 2) = 3$

When $p = -5$,
$m_2 = -5(-5 + 2) = 15$

3. Calculate the acute angle between the two tangents.

Let $\theta_1$ and $\theta_2$ be the angles the tangents make with the positive $x$-axis.
$\tan \theta_1 = 3 \implies \theta_1 = 71.565^\circ$
$\tan \theta_2 = 15 \implies \theta_2 = 86.186^\circ$

$\text{Acute angle} = 86.186^\circ - 71.565^\circ = 14.621^\circ \approx 14.6^\circ \text{ (1 d.p.)}$

(Alternatively, using the formula $\tan \alpha = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$, we get $\tan \alpha = \frac{12}{46} = \frac{6}{23} \implies \alpha \approx 14.6^\circ$)