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If $θ$ is the angle between the lines joining the origin to the points of intersection of the curve $2 x^{2} + 3 y^{2} = 6$ and the line $x + y = 1,$ then $\sin θ =$

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$\sqrt{\frac{96}{145}}$

$\sqrt{\frac{7}{145}}$

$\frac{1}{2}$

answer is C.

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Detailed Solution

$2 x^{2} + 3 y^{2} = 6 {(x + y)}^{2}$

$2 x^{2} + 3 y^{2} = 6 (x^{2} + y^{2} + 2 x y)$

$2 x^{2} + 3 y^{2} - 6 x^{2} - 6 y^{2} - 12 x y$

$4 x^{2} + 3 y^{2} + 12 x y = 0$

$\sin θ = \frac{2 \sqrt{h^{2} - a b}}{\sqrt{{(a - b)}^{2} + 4 h^{2}}} = \frac{2 \sqrt{36 - 12}}{\sqrt{1 + 4 \times 36}} \Rightarrow \frac{2 \sqrt{24}}{\sqrt{145}} = \sqrt{\frac{96}{145}}$

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