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(D) The equation of the line is $y - 3x = 2$,which can be written as $\frac{y - 3x}{2} = 1$.To homogenize the curve equation $x^2 + 2xy + 3y^2 + 4x +

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$	an^{-1}\left(\frac{2\sqrt{2}}{3}ight)$

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(D) The equation of the line is $y - 3x = 2$,which can be written as $\frac{y - 3x}{2} = 1$.To homogenize the curve equation $x^2 + 2xy + 3y^2 + 4x + 8y - 11 = 0$,we substitute $1$ with $\frac{y - 3x}{2}$:$x^2 + 2xy + 3y^2 + (4x + 8y)\left(\frac{y - 3x}{2}ight) - 11\left(\frac{y - 3x}{2}ight)^2 = 0$.Multiplying by $4$ to clear the denominator:$4x^2 + 8xy + 12y^2 + 2(4x + 8y)(y - 3x) - 11(y - 3x)^2 = 0$.$4x^2 + 8xy + 12y^2 + 2(4xy - 12x^2 + 8y^2 - 24xy) - 11(y^2 - 6xy + 9x^2) = 0$.$4x^2 + 8xy + 12y^2 - 24x^2 - 40xy + 16y^2 - 99x^2 + 66xy - 11y^2 = 0$.$-119x^2 + 34xy + 17y^2 = 0$,which simplifies to $17y^2 + 34xy - 119x^2 = 0$,or $y^2 + 2xy - 7x^2 = 0$.For a homogeneous equation $ax^2 + 2hxy + by^2 = 0$,the angle $	heta$ is given by $	an 	heta = \left|\frac{2\sqrt{h^2 - ab}}{a + b}ight|$.Here $a = -7, h = 1, b = 1$.$	an 	heta = \left|\frac{2\sqrt{1^2 - (-7)(1)}}{-7 + 1}ight| = \left|\frac{2\sqrt{8}}{-6}ight| = \frac{2(2\sqrt{2})}{6} = \frac{2\sqrt{2}}{3}$.Therefore,$	heta = 	an^{-1}\left(\frac{2\sqrt{2}}{3}ight)$.

The angle between the lines joining the origin to the points of intersection of the straight line $y = 3x + 2$ with the curve $x^2 + 2xy + 3y^2 + 4x + 8y - 11 = 0$ is

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