The square of the distance between the origin | vedclass

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Question

(A) The point of intersection $(x_0, y_0)$ of the lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by the solution of the partia

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$\frac{c(a + b) - f^2 - g^2}{ab - h^2}$

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(A) The point of intersection $(x_0, y_0)$ of the lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by the solution of the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.These are $2ax + 2hy + 2g = 0 \implies ax + hy + g = 0$ and $2hx + 2by + 2f = 0 \implies hx + by + f = 0$.Solving these equations using Cramer's rule,we get $x_0 = \frac{hf - bg}{ab - h^2}$ and $y_0 = \frac{gf - ah}{ab - h^2}$.The square of the distance from the origin $(0, 0)$ is $x_0^2 + y_0^2 = \frac{(hf - bg)^2 + (gf - ah)^2}{(ab - h^2)^2}$.Expanding the numerator: $h^2f^2 + b^2g^2 - 2hfbg + g^2f^2 + a^2h^2 - 2gfah = f^2(g^2 + h^2) + h^2(a^2 + b^2) - 2hfg(a + b)$.Using the condition for the pair of lines $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$,the expression simplifies to $\frac{c(a + b) - f^2 - g^2}{ab - h^2}$.

The square of the distance between the origin and the point of intersection of the lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is

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