JEE Main & Advanced Mathematics Pair of Straight Lines Question Bank Critical Thinking

  • question_answer Two of the lines represented by the equation \[a{{y}^{4}}+bx{{y}^{3}}+c{{x}^{2}}{{y}^{2}}+d{{x}^{3}}y+e{{x}^{4}}=0\] will be perpendicular, then [Kurukshetra CEE 1998]

    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)            \[(b-d)(ad-be)+{{(e-a)}^{2}}(a+c+e)=0\]

    D)            \[(b-d)(ad-be)+{{(e+a)}^{2}}(a+c+e)=0\]

    Correct Answer: A

    Solution :

               Let \[a{{y}^{4}}+bx{{y}^{3}}+c{{x}^{2}}{{y}^{2}}+d{{x}^{3}}y+e{{x}^{4}}\]                                 \[=(a{{x}^{2}}+pxy-a{{y}^{2}})({{x}^{2}}+qxy+{{y}^{2}})\]                    Comparing the coefficient of similar terms.                    We get, \[b=aq-p,\ c=-pq\], \[d=aq+p,\ e=-a\]                    \[b+d=2aq,\ e-a=-2a\]                     \[ad+be=2ap,a+c+e=-pq\]                    \[(b+d)(ad+be)=-{{(e-a)}^{2}}(a+c+e)\]                    \ \[(b+d)(ad+eb)+{{(e-a)}^{2}}(a+c+e)=0\].


You need to login to perform this action.
You will be redirected in 3 sec spinner