Which option is correct?

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2 Answers
May 14, 2017

All of them.

Explanation:

By inspection, all of the terms contain an #x# or #y# thus #(0,0)# is a solution for all of them for any a or b. Even though option 4 is only a point #(0,0)# It counts as a rational solution.

May 14, 2017

Option #1# and #3# only.

Explanation:

Expand:
#a^2y^2-2abxy+b^2x^2+4xy=0#
#(b^2)x^2+(4y-2aby)x+(a^2y^2)=0#

Using discriminant:
#(4y-2aby)^2-4(b^2)(a^2y^2)=n^2#

This simplifies to:
#16y^2(1-ab)=n^2#

For the equation to get rational solutions, then the equation should be equal a perfect square. #16y^2# is already one then #(1-ab)# should be too.

Option #1# and #3# only since they equal to #0# and #1/4# respectively.