If $θ i s$ the angle between the lines joining the origin to the points of intersection of the curve $2 x^{2} + 3 y^{2} = 6 a n d t h e l i n e x + y = 1, t h e n \sin θ =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

$\frac{1}{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 2 x^{2} + 3 y^{2} - 6 = 0 - - (1) \\ a n d l i n e x + y = 1 - - - (2) \\ H o m o g e n i z i n g (1) w i t h (2) \\ 2 x^{2} + 3 y^{2} - 6 {(x + y)}^{2} = 0 \\ \Rightarrow 2 x^{2} + 3 y^{2} - 6 (x^{2} + y^{2} + 2 x y) = 0 \\ \Rightarrow 2 x^{2} + 3 y^{2} - 6 x^{2} - 6 y^{2} - 12 x y = 0 \\ \Rightarrow - 4 x^{2} - 12 x y - 3 y^{2} = 0 \\ \Rightarrow 4 x^{2} + 12 x y + 3 y^{2} = 0 \\ C o m p a r i n g w i t h a x^{2} + 2 h x y + b y^{2} = 0 \\ a = 4, h = 6, b = 3 \\ \sin θ = \frac{2 \sqrt{h^{2} - a b}}{\sqrt{{(a - b)}^{2} + 4 h^{2}}} \\ = \frac{2 \sqrt{36 - 12}}{\sqrt{{(4 - 3)}^{2} + {(12)}^{2}}} = \frac{2 \sqrt{24}}{\sqrt{1 + 144}} = \sqrt{\frac{96}{145}} \end{array}$