Answer 1

1/2 

Given:
4y² + 2x − 20y + 17 = 0 

\[\Rightarrow y^2 + \frac{x}{2} - 5y + \frac{17}{4} = 0\]
\[ \Rightarrow \left( y - \frac{5}{2} \right)^2 + \frac{x}{2} - 2 = 0\]
\[ \Rightarrow \left( y - \frac{5}{2} \right)^2 = - 1\left( \frac{x}{2} - 2 \right)\]
\[ \Rightarrow \left( y - \frac{5}{2} \right)^2 = \frac{- 1}{2}\left( x - 4 \right)\] 

Let \[X = x - 4, Y = y - \frac{5}{2}\] 

\[\therefore Y^2 = \frac{- X}{2

∴ Length of the latus rectum = 4a =∴ Length of the latus rectum = 4a =}\]   

\[\frac{1}{2}\] units