A parabola is drawn with focus at \[\left( {3,4} \right)\] and vertex at the focus of another parabola ${y^2} - 12x - 4y + 4 = 0$. Then equation of the parabola is
A) ${x^2} - 6x - 8y + 25 = 0$
B) ${y^2} - 8x - 4y + 28 = 0$
C) ${x^2} - 6x + 8y - 25 = 0$
D) ${x^2} - 4x - 8y + 28 = 0$

Hint: First find the focus of the given parabola ${y^2} - 12x - 4y + 4 = 0$.
Also given that, the focus of the above parabola is the vertex of the asked parabola.
It is given that; A parabola is drawn with focus at \[\left( {3,4} \right)\] and vertex at the focus of another parabola ${y^2} - 12x - 4y + 4 = 0$.
Now, find the equation of the parabola by using formula ${\left( {x - h} \right)^2} = 4a\left( {y - k} \right)$.

Complete step by step solution:
A parabola is drawn with focus at \[\left( {3,4} \right)\] and vertex at the focus of another parabola ${y^2} - 12x - 4y + 4 = 0$.
Now, the equation ${y^2} - 12x - 4y + 4 = 0$ can be written as
 \[
  {y^2} - 4y + 4 = 12x \\
  {\left( {y - 2} \right)^2} = 12x
 \]
On comparing the above equation of parabola with ${\left( {y - h} \right)^2} = 4a\left( {x - k} \right)$ , we get \[h = {\text{2}},a = 3\] and \[k = 0\] .
So, the focus of the parabola will be \[\left( {a,h} \right) = \left( {3,2} \right)\] and vertex as \[\left( {k,h} \right) = \left( {0,2} \right)\]
Now, it is given that the parabola drawn has focus on the given parabola as its vertex.
So, the parabola which is to be drawn has focus \[\left( {3,4} \right)\] and vertex \[\left( {3,2} \right)\] .
So, its latus-rectum will be \[ = 4\left( 2 \right) = 8\] .
Thus, the equation of asked parabola can be given as
 $
  {\left( {x - 3} \right)^2} = 8\left( {y - 2} \right) \\
   \Rightarrow {x^2} - 6x + 9 = 8y - 16 \\
   \Rightarrow {x^2} - 6x - 8y + 25 = 0
 $

So, option (A) is correct.

Note:
Parabola:
A parabola is a U-shaped plane curve where any point is at an equal distance from a fixed point which is known as the focus and from a fixed straight line which is known as the directrix.