question_answer

If the lines \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] represents the adjacent sides of a parallelogram, then the equation of second diagonal if one is \[lx+my=1\], will be 

A)            \[(am+hl)x=(bl+hm)y\]   

B)            \[(am-hl)x=(bl-hm)y\]

C)            \[(am-hl)x=(bl+hm)y\]    

D)            None of these

Correct Answer: B

Solution :

           Let the equation of lines represented by \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] be \[y-{{m}_{1}}x=0\] and \[y-{{m}_{2}}x=0\] and one diagonal AC  be \[lx+my=1.\]            Therefore \[{{m}_{1}}+{{m}_{2}}=\frac{-2h}{b}\]and \[{{m}_{1}}{{m}_{2}}=\frac{a}{b}\]   Now on solving the equation of OA and OC with the line AC, we get the coordinates of            \[A\left( \frac{1}{l+m{{m}_{1}}},\frac{{{m}_{1}}}{l+m{{m}_{1}}} \right)\]            and \[C\left( \frac{1}{l+m{{m}_{2}}},\frac{{{m}_{2}}}{l+m{{m}_{2}}} \right)\]            Now find the coordinates of mid-point of AC and the equation of diagonal through this mid-point and origin. The required equation is \[x(am-hl)=(lb-mh)y\] .