Angle between the parabola $y^2=4(x-1)$ and $x^2+4(y-3)=0$ at the common end of their latus rectum is

Determine the angle between the parabolas at their intersection point.

The parabolas are $y^2 = 4(x-1)$ and $x^2 = -4(y-3)$, with their latus rectum endpoints determined by their respective focus coordinates.

For the parabola $y^2 = 4(x-1)$, the vertex is $(1,0)$ and the focus is $(2,0)$, giving latus rectum ends at $(2,2)$ and $(2,-2)$.

For $x^2 = -4(y-3)$, the vertex is $(0,3)$ and the focus is $(0,2)$; ends of the latus rectum correspond to $x = \pm 2$, $y=2$, resulting in endpoints of $(2,2)$ and $(-2,2)$.

The common point is $(2,2)$.

Differentiating $y^2 = 4(x-1)$ implicitly gives $2y\frac{dy}{dx} = 4$, so $\frac{dy}{dx} = \frac{2}{y}$. At $(2,2)$, the slope is $m_1 = 1$.

Differentiating $x^2 = -4(y-3)$ yields $2x = -4\frac{dy}{dx}$, so $\frac{dy}{dx} = -\frac{x}{2}$, giving $m_2 = -1$ at $(2,2)$.

The angle $\theta$ between the tangents follows from $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$, and with $m_1 = 1$, $m_2 = -1$, the denominator is $1 - 1 = 0$, indicating perpendicular tangents and $\theta = \frac{\pi}{2}$.

The angle between the parabolas at their common latus rectum end is $\frac{\pi}{2}$, corresponding to option A.