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(B) The equation of the pair of lines passing through the origin and the points of intersection of the two given curves is obtained by homogenizing th

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$g(a' + b') = g'(a + b)$

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(B) The equation of the pair of lines passing through the origin and the points of intersection of the two given curves is obtained by homogenizing the equations.Consider the curves:$S_1: ax^2 + 2hxy + by^2 + 2gx = 0$$S_2: a'x^2 + 2h'xy + b'y^2 + 2g'x = 0$To homogenize $S_1$ using $S_2$,we rewrite $S_1$ as $ax^2 + 2hxy + by^2 = -2gx$ and $S_2$ as $a'x^2 + 2h'xy + b'y^2 = -2g'x$.Multiplying $S_1$ by $g'$ and $S_2$ by $g$,we get:$g'(ax^2 + 2hxy + by^2) = -2gg'x$$g(a'x^2 + 2h'xy + b'y^2) = -2gg'x$Equating the two:$g'(ax^2 + 2hxy + by^2) = g(a'x^2 + 2h'xy + b'y^2)$$(ag' - a'g)x^2 + 2(hg' - h'g)xy + (bg' - b'g)y^2 = 0$For these lines to be mutually perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(ag' - a'g) + (bg' - b'g) = 0$$g'(a + b) - g(a' + b') = 0$$g(a' + b') = g'(a + b)$.

The lines joining the origin to the points of intersection of the curves $ax^2 + 2hxy + by^2 + 2gx = 0$ and $a'x^2 + 2h'xy + b'y^2 + 2g'x = 0$ will be mutually perpendicular,if

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