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(D) The equation of the line is $2x - 3y = 1$. We homogenize the equation of the curve $x^2 + 2xy + 5y^2 + 2x + 3y - 1 = 0$ using the line equation: $

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$\frac{3 \sqrt{2}}{7}$

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(D) The equation of the line is $2x - 3y = 1$. We homogenize the equation of the curve $x^2 + 2xy + 5y^2 + 2x + 3y - 1 = 0$ using the line equation: $x^2 + 2xy + 5y^2 + (2x + 3y)(1) - (1)^2 = 0$ Substituting $1 = 2x - 3y$: $x^2 + 2xy + 5y^2 + (2x + 3y)(2x - 3y) - (2x - 3y)^2 = 0$ $x^2 + 2xy + 5y^2 + (4x^2 - 9y^2) - (4x^2 - 12xy + 9y^2) = 0$ $x^2 + 2xy + 5y^2 + 4x^2 - 9y^2 - 4x^2 + 12xy - 9y^2 = 0$ $x^2 + 14xy - 13y^2 = 0$ This represents the pair of lines $OA$ and $OB$. Comparing with $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$2h = 14 \Rightarrow h = 7$,and $b = -13$. The angle $	heta$ between the lines is given by $	an 	heta = \frac{2\sqrt{h^2 - ab}}{|a + b|}$. $	an 	heta = \frac{2\sqrt{7^2 - (1)(-13)}}{|1 - 13|} = \frac{2\sqrt{49 + 13}}{12} = \frac{2\sqrt{62}}{12} = \frac{\sqrt{62}}{6}$. Using $\cos 	heta = \frac{1}{\sqrt{1 + 	an^2 	heta}} = \frac{1}{\sqrt{1 + \frac{62}{36}}} = \frac{1}{\sqrt{\frac{98}{36}}} = \frac{6}{\sqrt{98}} = \frac{6}{7\sqrt{2}} = \frac{3\sqrt{2}}{7}$.

Let the line $2x - 3y - 1 = 0$ intersect the curve $x^2 + 2xy + 5y^2 + 2x + 3y - 1 = 0$ at distinct points $A$ and $B$. If $O$ is the origin,then $\cos \angle AOB =$

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