The lines $3 x + 4 y - 5 = 0$ cuts the curve $2 x^{2} + 3 y^{2} = 5$ at A and B. If ‘O’ is the origin, then $∠ A O B =$

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$\frac{π}{3}$

$\frac{π}{6}$

$\frac{π}{8}$

$\frac{π}{2}$

answer is C.

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Detailed Solution

Given Equation $2 x^{2} + 3 y^{2} - 5 = 0 - - (1) a n d \frac{3 x + 4 y}{5} = 1 - - (2)$

Homogenizing (1) with (2)

$\begin{array}{l} 2 x^{2} + 3 y^{2} - 5 {(1)}^{2} = 0 \\ \Rightarrow 2 x^{2} + 3 y^{2} - 5 {(\frac{3 x + 4 y}{5})}^{2} = 0 \\ \Rightarrow 2 x^{2} + 3 y^{2} - 5 (\frac{9 x^{2} + 16 y^{2} + 24 x y}{25}) = 0 \\ \Rightarrow 10 x^{2} + 15 y^{2} - 9 x^{2} - 16 y^{2} - 24 x y = 0 \\ \Rightarrow x^{2} 24 x y - y^{2} = 0 \\ c o . e f f x^{2} + c o . e f f y^{2} = 1 - 1 = 0 \\ ∴ T h e l i n e s a r e ⊥ r \\ ∴ a n g l e b / w t h e l i n e s i s \frac{π}{2} \end{array}$