Q86 / 160 · TS EAMCET 2019 (03 May Shift 2) · Mathematics

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 $\sin θ =$

Solution

The homogenised equation of $2 x^{2} + 3 y^{2} = 6$ and $x + y = 1$ is

$2 x^{2} + 3 y^{2} - 6 (x + y)^{2} = 0$

$\Rightarrow - 4 x^{2} - 3 y^{2} - 12 x y = 0$

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

$\tan θ = |\frac{2 \sqrt{6^{2} - 12}}{7}|$

$= |\frac{4 \sqrt{6}}{7}|$

$\sin θ = \frac{4 \sqrt{6}}{\sqrt{145}} = \sqrt{\frac{96}{145}}$

Hence, the correct answer is option: 'C'.