Finding Equation of Parabola from given Focus and Vertex

1. Given a Parabola having Focus at a point \((x_f,y_f)\) and Vertex at a point \((x_v,y_v)\), the Co-efficients of Equation Directrix of the Parabola \(Ax + By + C=0\) can be found out using following steps
   1. Calculate the Projection Point of Focus/Vertex on the Directrix \((x_p,y_p)\). Since the Vertex of the Parabola is Mid Point Between the Focus and Projection Point of Focus/Vertex on the Directrix, the Projection Point can be calculated as
      
      \(\frac{\begin{bmatrix}x_f\\y_f\end{bmatrix}+\begin{bmatrix}x_p\\y_p\end{bmatrix}}{2}=\begin{bmatrix}x_v\\y_v\end{bmatrix}\)
      
      \(\Rightarrow \begin{bmatrix}x_p\\y_p\end{bmatrix}=2\begin{bmatrix}x_v\\y_v\end{bmatrix}-\begin{bmatrix}x_f\\y_f\end{bmatrix}\)   ...(1)
      
      
   2. Once the Projection Point of Focus/Vertex on the Directrix \((x_p,y_p)\) is calculated, the Co-efficients of Equation Directrix \(A_Dx + B_Dy + C_D=0\) can be calculated as follows
      
      \(A_D=x_v-x_f,\hspace{.3cm}B_D=y_v-y_f,\hspace{.3cm}C_D=-A_Dx_p-B_Dy_p\)   ...(2)
      
      
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 \(\mathbf{P}={A_D}^2 + {B_D}^2\)