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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: $2x^2-2xy+3y^2+(2x-y)(\frac{x+2y}{-\lambda}) - (\frac{x+2y}{-\lambda})^2 = 0$. Since the angle between these lines is $\frac{\pi}{2}$,the sum of the coefficients of $x^2$ and $y^2$ must be zero. Multiplying by $\lambda^2$,we get $2x^2\lambda^2-2xy\lambda^2+3y^2\lambda^2 - \lambda(2x^2+3xy-2y^2) - (x^2+4xy+4y^2) = 0$. The coefficient of $x^2$ is $2\lambda^2-2\lambda-1$ and the coefficient of $y^2$ is $3\lambda^2+2\lambda-4$. Setting their sum to zero: $(2\lambda^2-2\lambda-1) + (3\lambda^2+2\lambda-4) = 0$. $5\lambda^2-5=0$,which gives $\lambda^2=1$,so $\lambda = \pm 1$. Comparing with the given options,the correct value 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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