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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 $\mathbf{P}={A_D}^2 + {B_D}^2$