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(D) The given curve is $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$. The line is $x + 2y = k$,which implies $\frac{x + 2y}{k} = 1$. Homogenizing the curve equ

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(D) The given curve is $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$. The line is $x + 2y = k$,which implies $\frac{x + 2y}{k} = 1$. Homogenizing the curve equation using the line equation: $2x^2 - 2xy + 3y^2 + (2x - y)(1) - 1(1)^2 = 0$. Substituting $1 = \frac{x + 2y}{k}$: $2x^2 - 2xy + 3y^2 + (2x - y)\left(\frac{x + 2y}{k}\right) - \left(\frac{x + 2y}{k}\right)^2 = 0$. Multiplying by $k^2$: $k^2(2x^2 - 2xy + 3y^2) + k(2x^2 + 4xy - xy - 2y^2) - (x^2 + 4xy + 4y^2) = 0$. $x^2(2k^2 + 2k - 1) + xy(-2k^2 + 3k - 4) + y^2(3k^2 - 2k - 4) = 0$. Since the lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(2k^2 + 2k - 1) + (3k^2 - 2k - 4) = 0$. $5k^2 - 5 = 0$. $5k^2 = 5 \implies k^2 = 1$.

If the lines joining the origin to the points of intersection of the curve $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$ and the line $x + 2y = k$ are at right angles,then $k^2$ equals

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