Q. Two of the lines represented by the equation ${ay}^{4} + {bxy}^{3} + {cx}^{2}y^{2} + {dx}^{3}y + {ex}^{4} = 0$ will be perpendicular, then
A) $(b + d)(ad + be) + ({e - a)}^{2}(a + c + e) = 0$
B) $(b + d)(ad + be) + ({e + a)}^{2}(a + c + e) = 0$
C) $(b - d)(ad - be) + ({e - a)}^{2}(a + b + e) = 0$
D) $(b - d)(ad - be) + ({e + a)}^{2}(a + b + c) = 0$
Explanation
(a)
Let ${ay}^{4} + {bxy}^{3} + {cx}^{2}y^{2} + {dx}^{3}y + {ex}^{4} = \left( {ax}^{2} + pxy - {ay}^{2} \right)\left( x^{2} + qxy + y^{2} \right)$
On comparing the coefficient of similar terms, we get
$b = aq - p,\ c = - pq,\ d = aq + p,\ e = - a$
Now, $b + d = 2aq,\ \ \ e - a = - 2a$
$ad + be = 2ap,\ a + c + e = - pq$
$\therefore\ (b + d)(ad + be) = - (e - a)^{2}(a + c + e)$
$\Rightarrow \ (b + d)(ad + eb) + (e - a)^{2}(a + c + e) = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.