If the equation ax^{2}+2hxy+by^{2}+2gx+2fy=0 | vedclass

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(A) The equation of the pair of lines is $ax^{2}+2hxy+by^{2}+2gx+2fy=0$. Since one of the lines is the bisector of the angle between the coordinate ax

Vedclass · Accepted Answer

$(a+b)^{2}=4(h^{2}+g^{2})$

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(A) The equation of the pair of lines is $ax^{2}+2hxy+by^{2}+2gx+2fy=0$. Since one of the lines is the bisector of the angle between the coordinate axes,its equation is $y=x$ or $y=-x$,which can be written as $x-y=0$ or $x+y=0$. Let the other line be $lx+my+n=0$. Case $1$: If the line is $x-y=0$,then $(x-y)(lx+my+n) = lx^{2} + (m-l)xy - my^{2} + nx - ny = 0$. Comparing this with the given equation $ax^{2}+2hxy+by^{2}+2gx+2fy=0$,we get $a/l = 2h/(m-l) = b/(-m) = 2g/n = 2f/(-n)$. From $2g/n = 2f/(-n)$,we get $g = -f$. Also,$b/(-m) = 2f/(-n) \Rightarrow m = bn/(2f)$. Substituting these into the coefficients,we find the condition $(a+b)^{2} = 4(h^{2}+g^{2})$ or $4(h^{2}+f^{2})$. Given the options,the correct relation is $(a+b)^{2}=4(h^{2}+g^{2})$.

If the equation $ax^{2}+2hxy+by^{2}+2gx+2fy=0$ has one line as the bisector of the angle between the coordinate axes,then

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