If the equation ax^2 + 4xy - 2y^2 + 4x + 8y + | vedclass

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(B) The general equation of a second-degree curve is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. For this to represent a pair of straight lines, the con

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$(-\frac{3}{2}, \frac{1}{2})$

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(B) The general equation of a second-degree curve is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. For this to represent a pair of straight lines, the condition is $ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0$. Comparing coefficients: $A=a, H=2, B=-2, G=2, F=4, C=1$. Substituting these: $a(-2)(1) + 2(4)(2)(2) - a(4)^2 - (-2)(2)^2 - 1(2)^2 = 0 \implies -2a + 32 - 16a + 8 - 4 = 0 \implies -18a + 36 = 0 \implies a = 2$. The equation becomes $2x^2 + 4xy - 2y^2 + 4x + 8y + 1 = 0$. Let the point of intersection be $(h, k)$. The partial derivatives with respect to $x$ and $y$ are: $f_x = 4x + 4y + 4 = 0 \implies x + y = -1$ and $f_y = 4x - 4y + 8 = 0 \implies x - y = -2$. Adding the two equations: $2x = -3 \implies x = -\frac{3}{2}$. Subtracting: $2y = 1 \implies y = \frac{1}{2}$. The point of intersection is $(-\frac{3}{2}, \frac{1}{2})$.

If the equation $ax^2 + 4xy - 2y^2 + 4x + 8y + 1 = 0$ represents a pair of straight lines, then the coordinates of their point of intersection are...

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