Answer 1

Given: 
y² − 4y − 3x + 1 = 0 

\[\Rightarrow \left( y - 2 \right)^2 - 4 - 3x + 1 = 0\]
\[ \Rightarrow \left( y - 2 \right)^2 = 3\left( x + 1 \right)\]
\[ \Rightarrow \left( y - 2 \right)^2 = 3\left( x - \left( - 1 \right) \right)\] 

Let \[Y = y - 2\] 

\[X = x + 1\] 

Then, we have: 

\[Y^2 = 3X\] 

Comparing the given equation with\[Y^2 = 4aX\] 

\[4a = 3 \Rightarrow a = \frac{3}{4}\] 

∴ Vertex = (X = 0, Y = 0) = \[\left( x = - 1, y = 2 \right)\] 

Focus = (X = a, Y = 0) = \[\left( x + 1 = \frac{3}{4}, y - 2 = 0 \right) = \left( x = \frac{- 1}{4}, y = 2 \right)\]

Equation of the directrix:
X = −a
i.e. \[x + 1 = \frac{- 3}{4} \Rightarrow x = \frac{- 7}{4}\] 

Axis = Y = 0
i.e. \[y - 2 = 0 \Rightarrow y = 2\] 

Length of the latus rectum = 4a = 3 units