The point of concurrence of all the chords of | vedclass

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(B) Let the equation of the chord be $lx + my = 1$. Homogenizing the curve $3x^2 - y^2 - 2x + 4y = 0$ with the chord equation: $3x^2 - y^2 - 2x(lx + m

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$(1, -2)$

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(B) Let the equation of the chord be $lx + my = 1$. Homogenizing the curve $3x^2 - y^2 - 2x + 4y = 0$ with the chord equation: $3x^2 - y^2 - 2x(lx + my) + 4y(lx + my) = 0$ $3x^2 - y^2 - 2lx^2 - 2mxy + 4lxy + 4my^2 = 0$ $(3 - 2l)x^2 + (4l - 2m)xy + (4m - 1)y^2 = 0$ Since the chord subtends a right angle at the origin,the pair of lines represented by the homogeneous equation must be perpendicular. Therefore,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(3 - 2l) + (4m - 1) = 0$ $2 - 2l + 4m = 0$ $l - 2m = 1$ Comparing $lx + my = 1$ with $l(1) + m(-2) = 1$,we see that the line always passes through the point $(1, -2)$.

The point of concurrence of all the chords of the curve $3x^2 - y^2 - 2x + 4y = 0$ which subtend a right angle at the origin is

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