The equation of lines joining (0, 0) and points of intersection $x^{2} + y^{2} + 2 x y = 4, 3 x^{2} + 5 y^{2} - x y = 7$ is

x^{2} + y^{2} - x y = 0

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5 x^{2} + 2 y^{2} - 13 x y = 0

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5 x^{2} + y^{2} - 6 x y = 0

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5 x^{2} + 13 y^{2} - 18 x y = 0

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Solution

The correct option is D $5 x^{2} + 13 y^{2} - 18 x y = 0$

$C_{1} : x^{2} + y^{2} = 4 - 2 x y$

$\Rightarrow (x + y)^{2} = 4$
$\Rightarrow x + y = (\pm 2)$

$\Rightarrow (\frac{x + y}{\pm 2}) = 1$

$Now, By homogenisation$

$3 x^{2} + 5 y^{2} - x y = 7 (1)^{2}$

$\Rightarrow 3 x^{2} + 5 y^{2} - x y = 7 {(\frac{x + y}{\pm 2})}^{2}$

$\Rightarrow 3 x^{2} + 5 y^{2} - x y = \frac{7}{4} (x^{2} + y^{2} + 2 x y)$

$\Rightarrow 12 x^{2} + 20 y^{2} - 4 x y = 7 x^{2} + 7 y^{2} + 14 x y$
$\Rightarrow 5 x^{2} + 13 y^{2} - 18 x y = 0$

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