Finding Equation of Hyperbola from a given Directrix, Adjacent Focus and Eccentricity

1. Given the Equation of a Directrix \(Ax + By + C=0\), Coordinates of it's Adjacent Focus \((x_{f1},y_{f1})\) and Eccentricity \(e\) the following gives the steps for calculation of the Equation of the Hyperbola
2. Calculate the Signed Distance \(d\) Between the given Focus and the Directrix as follows
   
   \(d=\frac{Ax_{f1} + By_{f1} + C}{\sqrt{A^2 +B^2}}\)   ...(1)
3. Calculate the Coordinates of Projection of the Focus on the Directrix as follows
   
   \(\begin{bmatrix}x_p\\y_p\end{bmatrix}=\begin{bmatrix}x_{f1}\\y_{f1}\end{bmatrix} - d\begin{bmatrix}\frac{A}{\sqrt{A^2 +B^2}}\\\frac{B}{\sqrt{A^2 +B^2}}\end{bmatrix}\)   ...(2)
4. Calculate the Coordinates of the Vertex \((x_{v1},y_{v1})\) lying between Focus and Directrix as follows
   
   We know that the Coordinates of Vertex \((x_{v1},y_{v1})\) Divides the Line Joining the Coordinates of Focus \((x_{f1},y_{f1})\) and Coordinates of Projection of the Focus on the Directrix \((x_p,y_p)\) intenally in a Ratio \(e:1\). Therefore using Section Formula we have
   
   \(\begin{bmatrix}x_{v1}\\y_{v1}\end{bmatrix}=\begin{bmatrix}\frac{x_{f1} + ex_p}{e+1}\\\frac{y_{f1} + ey_p}{e+1}\end{bmatrix}\)   ...(3)
5. Once we get Coordinates of Both the Adjacent Focus and the Vertex the Equation of the Hyperbola can be determined as given in the topic Finding Equation of Hyperbola from a given Focus, a Vertex and Eccentricity.