Finding Equation of Parabola from given Vertex and Directrix

1. Given a Parabola having Vertex at a point \((x_v,y_v)\) and Equation Directrix as \(A_Dx + B_Dy + C_D=0\), it's Coordinates of Focus \((x_f,y_f)\) can be found out using following steps
   1. Calculate the Signed Distance between the Vertex and the Directrix as follows
      
      \(D_{VD}=\frac{A_Dx_v + B_Dy_v + C_D}{\sqrt{{A_D}^2 + {B_D}^2}}\)   ...(1)
   2. Calculate the Projection Point of Vertex on the Directrix \((x_p,y_p)\) as follows
      
      \(\begin{bmatrix}x_p\\y_p\end{bmatrix}=\begin{bmatrix}x_v\\y_v\end{bmatrix} - D_{VD}\begin{bmatrix}\frac{A_D}{\sqrt{{A_D}^2 + {B_D}^2}}\\\frac{B_D}{\sqrt{{A_D}^2 + {B_D}^2}}\end{bmatrix}\)   ...(2)
   3. Since the Vertex of the Parabola is Mid Point Between the Focus and Projection Point of Vertex on the Directrix, the Coordinates of Focus \((x_f,y_f)\) can be calculated as
      
      \(\frac{\begin{bmatrix}x_f\\y_f\end{bmatrix}+\begin{bmatrix}x_p\\y_p\end{bmatrix}}{2}=\begin{bmatrix}x_v\\y_v\end{bmatrix}\)
      
      \(\Rightarrow \begin{bmatrix}x_f\\y_f\end{bmatrix}=2\begin{bmatrix}x_v\\y_v\end{bmatrix}-\begin{bmatrix}x_p\\y_p\end{bmatrix}\)   ...(3)
      
      
2. Now, as given in Derivation and Properties of Implicit Coordinate Equation for Axis Aligned and Arbitrarily Rotated Parabolas, for a Parabola having it's Focus at a point \((x_f,y_f)\) and having Directrix given by the equation \(A_Dx + B_Dy + C_D=0\), the Equation of Parabola is given as
   
   \({B_D}^2x^2 - 2A_DB_Dxy + {A_D}^2y^2 - 2(Px_f + A_DC_D)x - 2(Py_f + B_DC_D)y + P{x_f}^2 + P{y_f}^2 - {C_D}^2 =0\)   ...(1)
   
   where \(\mathbf{P}={A_D}^2 + {B_D}^2\)