Consider the parabola $25\left[(x-2)^2+(y+5)^2\right]=(3 x+4 y-1)^2$, match the characteristic of this parabola given in List I with its corresponding item in List II.

$$ \begin{array}{lll} \hline & \text { List I } & \text { List II } \\\\ \hline \text { I } & \text { Vertex } & \text { (A) } 8 \\\\ \hline \text { II } & \text { length of latus rectum } & \text { (B) }\left(\frac{29}{10}, \frac{-38}{10}\right) \\\\ \hline \text { III } & \text { Directrix } & \text { (C) } 3 x+4 y-1=0 \\\\ \hline \text { IV } & \begin{array}{l} \text { One end of the latus } \\\\ \text { rectum } \end{array} & \text { (D) }\left(\frac{-2}{5}, \frac{-16}{5}\right) \\\\ \hline \end{array} $$

The correct answer is

If $\mathbf{A B}$ is the focal chord of the parabola $y^2=16 x$ and $A=(1,-4)$, then the equation of the normal to the parabola at the point $B$ is

If one of the vertices of an equilateral triangle inscribed in the parabola $y^2=12 x$ coincides with the vertex of the parabola, then the area (in sq units) of that triangle is