A chord through the point (1,-2) cuts the cur | vedclass

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(D) The given curve is $3x^2-y^2-2x+4y=0$. Let the chord passing through $(1,-2)$ have slope $m$. The equation of the line is $y+2=m(x-1)$,which can b

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(D) The given curve is $3x^2-y^2-2x+4y=0$. Let the chord passing through $(1,-2)$ have slope $m$. The equation of the line is $y+2=m(x-1)$,which can be written as $\frac{mx-y}{m+2}=1$. To find the joint equation of the lines $OP$ and $OQ$,we homogenize the curve equation using the line equation: $3x^2-y^2-(2x-4y)\left(\frac{mx-y}{m+2}ight)=0$. Multiplying by $(m+2)$,we get: $(m+2)(3x^2-y^2)-(2x-4y)(mx-y)=0$. Expanding this: $3mx^2+6x^2-my^2-2y^2-(2mx^2-2xy-4mxy+4y^2)=0$. Grouping terms: $x^2(3m+6-2m) + xy(2+4m) + y^2(-m-2-4) = 0$. $x^2(m+6) + xy(4m+2) + y^2(-m-6) = 0$. The equation is of the form $ax^2+2hxy+by^2=0$. Here,$a = m+6$ and $b = -(m+6)$. Since $a+b = (m+6) - (m+6) = 0$,the lines $OP$ and $OQ$ are perpendicular. Therefore,$	heta = 90^{\circ}$.

$A$ chord through the point $(1,-2)$ cuts the curve $3x^2-y^2-2x+4y=0$ at $P$ and $Q$. If $PQ$ subtends an angle $\theta$ at the origin,then $\theta$ equals (in $^{\circ}$)

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