• # 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)            ${{f}^{2}}+{{g}^{2}}=a{{f}^{2}}+b{{g}^{2}}$

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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