Question

Write the distance between the vertex and focus of the parabola y² + 6y + 2x + 5 = 0. 

Answer 1

Given: 

\[y^2 + 6y + 2x + 5 = 0\] 

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

Let Y = y+3,  \[X = x - 2\]

From (1), we have: 

\[Y^2 = - 2X\] 

Putting \[4a = 2\] 

\[a = \frac{1}{2}\] 

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

Vertex = \[\left( X = 0, Y = 0 \right) = \left( x = 2, y = - 3 \right)\] 

Thus, we have:
Focus =\[\left( \frac{3}{2}, - 3 \right)\] 

Vertex = \[\left( 2, - 3 \right)\] 

 Distance between the vertex and the focus:

\[\sqrt{\left( \frac{3}{2} - 2 \right)^2 + \left( - 3 + 3 \right)^2}\]
\[\sqrt{\left( \frac{1}{2} \right)^2}\]
\[ = \frac{1}{2} \text{ units }\]