How to strictly grep (with before and after args) for last N number of occurrences?

Shell Programming and Scripting
1

Here is my grep string to print only the last occurrence, with before and after lines. Note that the tail Argument is sum of B and A args + 1, so that it prints the data of only the last 1 match. Now I need to print last 2 such matches. I thought of doubling the tail arg like 5+5+1 (For -- line), for ex: tail -11 should return last 2 matches, but I want it to be strict (I guess its tail -maxNumber, but I want it to be something like tail -MinNumber), so it has to either print 2 occurrences or nothing. Right now even if there is only one sample, it returns the output, I want output only if there are at least 2 matches. May be we can pipe it to awk to solve this. Please help.

grep -h -B 1 -A 3 'Authentication Error' MyLogFile.txt | tail -5

Output is:

	line1 Content
	Authentication Error while calling
	line3 Content
	line4 Content
	line5 Content

If there are multiple occurrences:

grep -h -B 1 -A 3 'Authentication Error' MyLogFile.txt | tail -11

Output is:

	line1 Content
	Authentication Error while calling
	line3 Content
	line4 Content
	line5 Content
--	  
	line11 Content
	Authentication Error while calling
	line13 Content
	line14 Content
	line15 Content

But even if there is only one occurrence:

grep -h -B 1 -A 3 'Authentication Error' MyLogFile.txt | tail -11

Output is:

	line1 Content
	Authentication Error while calling
	line3 Content
	line4 Content
	line5 Content
2

There's no such option. You may check the file twice like this:

[ "$(grep -c "Authentication Error" MyLogFile.txt)" -ge 2 ] && \
   grep -h -B 1 -A 3 'Authentication Error' MyLogFile.txt | tail -11
3

Thank you so much! It works! In fact I am open for complete alternative with sed or awk too, provided it lets me configure before and after lines like this one.