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

1. Given the Equation of a Directrix \(Ax + By + C=0\), Coordinates of it's Adjacent Vertex \((x_{v1},y_{v1})\) 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 Vertex and the Directrix as follows
   
   \(d=\frac{Ax_{v1} + By_{v1} + C}{\sqrt{A^2 +B^2}}\)   ...(1)
3. Calculate the Coordinates of Projection of the Vertex on the Directrix as follows
   
   \(\begin{bmatrix}x_p\\y_p\end{bmatrix}=\begin{bmatrix}x_{v1}\\y_{v1}\end{bmatrix} - d\begin{bmatrix}\frac{A}{\sqrt{A^2 +B^2}}\\\frac{B}{\sqrt{A^2 +B^2}}\end{bmatrix}\)   ...(2)
4. Calculate the Coordinates of the Focus \((x_{f1},y_{f1})\) Adjacent to given Vertex 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) \(\Rightarrow \begin{bmatrix}x_{f1}\\ y_{f1} \end{bmatrix}=\begin{bmatrix}x_{v1} (e+1) - ex_p \\y_{v1} (e+1) - ey_p \end{bmatrix}\)   ...(4)
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.