Equation of the pair of lines joining the origin to the points of intersection of the line and the conic

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Solution

A) $2 x + 3 y - 1 = 0 \Rightarrow 2 x + 3 y = 1$ ...(1)
$x^{2} + y^{2} = 5 \Rightarrow x^{2} + y^{2} = 5 {(1)}^{2}$
Substituting value of from (1), we get
$x^{2} + y^{2} = 5 {(2 x + 3 y)}^{2} \Rightarrow 19 x^{2} + 60 x y + 44 y^{2} = 0$
B) $x + y + 1 = 0 \Rightarrow (x + y) = - 1$ ...(2)
$x^{2} + y^{2} - 2 x + 2 y + 1 = 0 \Rightarrow x^{2} + y^{2} - 2 (x - y) (- 1) + 1 {(- 1)}^{2} = 0$
Substituting value of -1 from (2), we get
$x^{2} + y^{2} - 2 (x - y) (x + y) + (x + y)^{2} = 0 \Rightarrow x y + 2 y^{2} = 0$
C) $x - y = 1$ ...(3)
$x y = 4 \Rightarrow x y = 4 {(1)}^{2}$
Substituting value of 1 from (3) we get
$x y = 4 (x - y)^{2} \Rightarrow 4 x^{2} - 9 x y + 4 y^{2} = 0$
D) $x = α \Rightarrow \frac{x}{α} = 1$ ...(4)
$y^{2} = 4 α x \Rightarrow y^{2} = 4 α x (1)$
Substituting value of 1 from (4), we get
$y^{2} = 4 a x (\frac{x}{α}) \Rightarrow 4 x^{2} - y^{2} = 0$

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