求 $p$ 的值，使得直线 $y = x + 1$ 是曲线 $y = x^2 + (2p + 3)x + p + 4$ 的切线。 利用第 (i) 部分中求得的 $p$ 值，求直线与曲线相交点的坐标。

1. Equate the equations of the line and the curve to find their points of intersection.

$x^2 + (2p + 3)x + p + 4 = x + 1$

2. Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$.

$x^2 + (2p + 3 - 1)x + p + 4 - 1 = 0$
$x^2 + (2p + 2)x + p + 3 = 0$

3. Apply the condition for a line to be a tangent to a curve (discriminant equals zero).

$b^2 - 4ac = 0$
$(2p + 2)^2 - 4(1)(p + 3) = 0$

4. Solve the resulting quadratic equation for $p$.

$4p^2 + 8p + 4 - 4p - 12 = 0$
$4p^2 + 4p - 8 = 0$
$p^2 + p - 2 = 0$
$(p + 2)(p - 1) = 0$
$p = -2 \text{ or } p = 1$

1. State the values of $p$ from part (i) and the intersection equation

From part (i), the values of $p$ are $p = -2$ and $p = 1$.
The intersection equation is $x^2 + (2p + 2)x + p + 3 = 0$.

2. Find the coordinates for $p = -2$

Substitute $p = -2$ into the intersection equation:
$x^2 + (2(-2) + 2)x + (-2) + 3 = 0$
$x^2 - 2x + 1 = 0$
$(x - 1)^2 = 0$
$x = 1$
Substitute $x = 1$ into the line equation $y = x + 1$:
$y = 1 + 1 = 2$
The point is $(1, 2)$.

3. Find the coordinates for $p = 1$

Substitute $p = 1$ into the intersection equation:
$x^2 + (2(1) + 2)x + 1 + 3 = 0$
$x^2 + 4x + 4 = 0$
$(x + 2)^2 = 0$
$x = -2$
Substitute $x = -2$ into the line equation $y = x + 1$:
$y = -2 + 1 = -1$
The point is $(-2, -1)$.