The lines joining the points of intersection | vedclass

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(C) To find the lines joining the origin to the points of intersection,we homogenize the equation of the curve using the line equation $x - y = 2$,whi

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(C) To find the lines joining the origin to the points of intersection,we homogenize the equation of the curve using the line equation $x - y = 2$,which implies $\frac{x - y}{2} = 1$.The equation of the curve is $5x^2 + 12xy - 8y^2 + 8x - 4y + 12 = 0$.Substituting $1$ with $\frac{x - y}{2}$ in the linear and constant terms:$5x^2 + 12xy - 8y^2 + (8x - 4y)\left(\frac{x - y}{2}\right) + 12\left(\frac{x - y}{2}\right)^2 = 0$Expanding the terms:$5x^2 + 12xy - 8y^2 + (4x^2 - 4xy - 2xy + 2y^2) + 12\left(\frac{x^2 - 2xy + y^2}{4}\right) = 0$$5x^2 + 12xy - 8y^2 + 4x^2 - 6xy + 2y^2 + 3(x^2 - 2xy + y^2) = 0$$5x^2 + 12xy - 8y^2 + 4x^2 - 6xy + 2y^2 + 3x^2 - 6xy + 3y^2 = 0$Combining like terms:$(5 + 4 + 3)x^2 + (12 - 6 - 6)xy + (-8 + 2 + 3)y^2 = 0$$12x^2 + 0xy - 3y^2 = 0$$12x^2 - 3y^2 = 0$$4x^2 - y^2 = 0$This factors as $(2x - y)(2x + y) = 0$,so the lines are $y = 2x$ and $y = -2x$.Since the slopes are $m_1 = 2$ and $m_2 = -2$,the lines are symmetric with respect to the axes. The angles made with the axes are equal in magnitude.

The lines joining the points of intersection of the curve $5x^2 + 12xy - 8y^2 + 8x - 4y + 12 = 0$ and the line $x - y = 2$ to the origin make angles with the axes that are:

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