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(A) The equation of the pair of lines joining the origin to the points of intersection of the line $x+2y+\lambda=0$ and the curve $2x^2-2xy+3y^2+2x-y-

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$1$

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(A) The equation of the pair of lines joining the origin to the points of intersection of the line $x+2y+\lambda=0$ and the curve $2x^2-2xy+3y^2+2x-y-1=0$ is obtained by homogenizing the curve equation using the line equation: $2x^2-2xy+3y^2+(2x-y)(\frac{x+2y}{-\lambda}) - (\frac{x+2y}{-\lambda})^2 = 0$. Multiplying by $\lambda^2$,we get: $\lambda^2(2x^2-2xy+3y^2) - \lambda(2x-y)(x+2y) - (x+2y)^2 = 0$. Expanding this,the coefficient of $x^2$ is $2\lambda^2 - 2\lambda - 1$ and the coefficient of $y^2$ is $3\lambda^2 + 2\lambda - 4$. Since the angle between the lines is $\frac{\pi}{2}$,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(2\lambda^2 - 2\lambda - 1) + (3\lambda^2 + 2\lambda - 4) = 0$. $5\lambda^2 - 5 = 0 \implies \lambda^2 = 1 \implies \lambda = \pm 1$. Thus,the value of $\lambda$ is $1$.

If the angle between the lines joining the origin to the points of intersection of $x+2y+\lambda=0$ and $2x^2-2xy+3y^2+2x-y-1=0$ is $\frac{\pi}{2}$,then a value of $\lambda$ is

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