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

  • question_answer The lines represented by the equation \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\] will be equidistant from the origin, if

    A)            \[{{f}^{2}}+{{g}^{2}}=c(b-a)\]                                           

    B)            \[{{f}^{4}}+{{g}^{4}}=c(b{{f}^{2}}+a{{g}^{2}})\]

    C)            \[{{f}^{4}}-{{g}^{4}}=c(b{{f}^{2}}-a{{g}^{2}})\]          

    D)            \[{{f}^{2}}+{{g}^{2}}=a{{f}^{2}}+b{{g}^{2}}\]

    Correct Answer: C

    Solution :

               Let the equations represented by            \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\] be            \[lx+my+n=0\] and \[l'x+m'y+n'=0\].            Then the combined equation represented by these lines is given by \[(lx+my+n)(l'x+m'y+n')=0\]            So, it must be similar with the given equation.            On comparing, we get            \[ll'=a,\,\ mm'=b\ \ nn'=c,\ \ lm'+ml'=2h,\ \ ln'+l'n=2g\],            \[mn'+nm'=2f\]            According to the condition, the length of perpendiculars drawn from origin to the lines are same.            So, \[\frac{n}{\sqrt{{{l}^{2}}+{{m}^{2}}}}=\frac{n'}{\sqrt{l{{'}^{2}}+m{{'}^{2}}}}=\frac{{{(nn')}^{2}}}{({{l}^{2}}+{{m}^{2}})(l{{'}^{2}}+m{{'}^{2}})}\]            Now on eliminating \[l,\ m,\ l',\ m'\]and \[n,\ n'\], we get the required condition \[{{f}^{4}}-{{g}^{4}}=c(b{{f}^{2}}-a{{g}^{2}}).\]

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