Question

Find the vertex, focus, axis, directrix and latus-rectum of the following parabola 

x² + y = 6x − 14

Answer 1

Given:
x² = 6x−y−14 

\[\Rightarrow \left( x - 3 \right)^2 = - y - 14 + 9\]
\[ \Rightarrow \left( x - 3 \right)^2 = - y - 5 = - \left( y + 5 \right)\] 

Let\[Y = y + 5\] \[X = x - 3\] 

Then, we have:

\[X^2 = - Y\]

Comparing the given equation with\[X^2 = - 4aY\] 

\[4a = 1 \Rightarrow a = \frac{1}{4}\] 

∴ Vertex = (X = 0, Y = 0) = \[\left( x = 3, y = - 5 \right)\] 

Focus = (X = 0, Y = −a) = \[\left( x - 3 = 0, y + 5 = \frac{- 1}{4} \right) = \left( x = 3, y = \frac{- 21}{4} \right)\]

Equation of the directrix:
Y = a
i.e.\[y + 5 = \frac{1}{4} \Rightarrow y = \frac{- 19}{4}\] 

Axis = X = 0
i.e \[x - 3 = 0 \Rightarrow x = 3\]

Length of the latus rectum = 4a = 1 units