If the pair of lines ax^2 + 2hxy + by^2 + 2gx | vedclass

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(A) The given equation of the pair of lines is $f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. To find the point of intersection,we take partial d

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$2fgh = bg^2 + ch^2$

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(A) The given equation of the pair of lines is $f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. To find the point of intersection,we take partial derivatives with respect to $x$ and $y$: $\frac{\partial f}{\partial x} = 2ax + 2hy + 2g = 0 \implies ax + hy + g = 0$ $\frac{\partial f}{\partial y} = 2hx + 2by + 2f = 0 \implies hx + by + f = 0$ Since the lines intersect on the $y$-axis,the $x$-coordinate of the point of intersection is $x = 0$. Substituting $x = 0$ into the first partial derivative equation: $h(y) + g = 0 \implies y = -g/h$. Now,substitute $(0, -g/h)$ into the original equation $f(x, y) = 0$: $a(0)^2 + 2h(0)(-g/h) + b(-g/h)^2 + 2g(0) + 2f(-g/h) + c = 0$ $b(g^2/h^2) - 2fg/h + c = 0$ Multiplying by $h^2$,we get $bg^2 - 2fgh + ch^2 = 0$,which simplifies to $bg^2 + ch^2 = 2fgh$.

If the pair of lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ intersect on the $y$-axis,then:

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