If the chord through (1, -2) cuts the curve 3 | vedclass

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(D) Let $PQ$ subtend the angle $	heta$ at the origin,so $\angle POQ = 	heta$.Let the slope of the line be $m$.The equation of the line passing throu

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$90$

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(D) Let $PQ$ subtend the angle $	heta$ at the origin,so $\angle POQ = 	heta$.Let the slope of the line be $m$.The equation of the line passing through $(1, -2)$ is $y + 2 = m(x - 1)$.Rearranging,we get $mx - y = m + 2$,or $\frac{mx - y}{m + 2} = 1$  ...$(i)$The given curve is $3x^2 - y^2 - 2x + 4y = 0$  ...(ii)To find the angle subtended at the origin,we homogenize the equation of the curve using the line equation:$3x^2 - y^2 - 2x(1) + 4y(1) = 0$Substituting $1 = \frac{mx - y}{m + 2}$:$3x^2 - y^2 - 2x\left(\frac{mx - y}{m + 2}ight) + 4y\left(\frac{mx - y}{m + 2}ight) = 0$Multiplying by $(m + 2)$:$3(m + 2)x^2 - (m + 2)y^2 - 2x(mx - y) + 4y(mx - y) = 0$$(3m + 6)x^2 - (m + 2)y^2 - 2mx^2 + 2xy + 4mxy - 4y^2 = 0$$(m + 6)x^2 + (4m + 2)xy - (m + 6)y^2 = 0$For the angle subtended to be $90^{\circ}$,the sum of the coefficients of $x^2$ and $y^2$ must be zero.Coefficient of $x^2$ + Coefficient of $y^2 = (m + 6) + (-(m + 6)) = 0$.Since the sum is zero,the angle subtended by $PQ$ at the origin is $90^{\circ}$.

If the chord through $(1, -2)$ cuts the curve $3x^2 - y^2 - 2x + 4y = 0$ at $P$ and $Q$,then the angle subtended by $PQ$ at the origin is (in $^{\circ}$)

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