The y-axis and the lines (a5 – 2a3 )x + (a + 2)y + 3a = 0 and (a5 – 3a2 )x + 4y + a – 2 = 0 are concurrent for
Two values of a
Three values of a
Five values of a
no value of a
Let the lines intersect at the point k then
(a+2)k+3a=0 and 4k+a−2=0
⇒3aa+2=a−24⇒a2−12a+4=0
which gives us two values of a