JEE Main2012MathematicsStraight LinesActual
The point of intersection of the lines (a^3+3 ) x+a y+a-3=0 and (a^5+2 ) x+(a+2) y+2 a+3=0 (a real) lies on the y -axis for
Options
- Ano value of a
- Bmore than two values of a
- Cexactly one value of a
- Dexactly two values of a
Correct answer
A. no value of a
Step-by-step solution
Given equation of lines are (a^3+3 ) x+a y+a-3=0 and (a^5+2 ) x+(a+2) y+2 a+3=0 (a real) Since point of intersection of lines lies on y -axis. Put x=0 in each equation, we get a y +a-3=0 and (a+2) y+2 a+3=0 On solving these we get aligned & (a+2)(a-3)-a(2 a+3)=0 & a^2-a-6-2 a^2-3 a=0 & -a^2-4 a-6=0 a^2+4 a+6=0 & a= -4 16-24 2 = -4 -8 2 & ( not real) aligned This shows that the point of intersection of the lines lies on the y -axis for no value of ' a '.