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(C) The equation of the curve is $2x^2 + 3y^2 = 6$,which can be written as $2x^2 + 3y^2 - 6(1)^2 = 0$. Substituting $1 = x + y$ from the line equation

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The equation of the curve is $2x^2 + 3y^2 = 6$,which can be written as $2x^2 + 3y^2 - 6(1)^2 = 0$. Substituting $1 = x + y$ from the line equation,we get the homogeneous equation of the pair of lines passing through the origin: $2x^2 + 3y^2 - 6(x + y)^2 = 0$ $2x^2 + 3y^2 - 6(x^2 + y^2 + 2xy) = 0$ $-4x^2 - 3y^2 - 12xy = 0$ $4x^2 + 12xy + 3y^2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 = 0$,we have $a = 4$,$2h = 12$ (so $h = 6$),and $b = 3$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$. $	an 	heta = \left| \frac{2\sqrt{6^2 - 4(3)}}{4 + 3} ight| = \left| \frac{2\sqrt{36 - 12}}{7} ight| = \frac{2\sqrt{24}}{7} = \frac{4\sqrt{6}}{7}$. Since $	an 	heta = \frac{4\sqrt{6}}{7}$,we construct a right triangle with opposite side $4\sqrt{6}$ and adjacent side $7$. The hypotenuse is $\sqrt{(4\sqrt{6})^2 + 7^2} = \sqrt{16 	imes 6 + 49} = \sqrt{96 + 49} = \sqrt{145}$. Therefore,$\sin 	heta = \frac{4\sqrt{6}}{\sqrt{145}} = \sqrt{\frac{16 	imes 6}{145}} = \sqrt{\frac{96}{145}}$.

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

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