If the equation 3x^2+7xy+2y^2+2gx+2fy+2=0 rep | vedclass

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(D) The general equation of a second-degree curve $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines if $abc+2fgh-af^2-bg^2-ch^2=0$.For

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$\frac{25}{2}$

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(D) The general equation of a second-degree curve $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines if $abc+2fgh-af^2-bg^2-ch^2=0$.For the given equation $3x^2+7xy+2y^2+2gx+2fy+2=0$,we have $a=3, h=\frac{7}{2}, b=2, g=g, f=f, c=2$.Substituting these values into the condition:$3(2)(2) + 2f g(\frac{7}{2}) - 3f^2 - 2g^2 - 2(\frac{7}{2})^2 = 0$$12 + 7fg - 3f^2 - 2g^2 - \frac{49}{2} = 0$$24 + 14fg - 6f^2 - 4g^2 - 49 = 0$$6f^2 + 4g^2 - 14fg = -25$  ---$(i)$To find the point of intersection $(x, y)$,we partially differentiate the equation with respect to $x$ and $y$:$\frac{\partial}{\partial x}(3x^2+7xy+2y^2+2gx+2fy+2) = 6x+7y+2g = 0$  ---(ii)$\frac{\partial}{\partial y}(3x^2+7xy+2y^2+2gx+2fy+2) = 7x+4y+2f = 0$  ---(iii)Solving (ii) and (iii) for $x$ and $y$ using Cramer's rule:$x = \frac{-(2g)(4) - (-2f)(7)}{(6)(4) - (7)(7)} = \frac{-8g+14f}{-25} = \frac{8g-14f}{25}$$y = \frac{(6)(-2f) - (7)(-2g)}{-25} = \frac{-12f+14g}{-25} = \frac{12f-14g}{25}$The square of the distance from the origin is $x^2+y^2 = \frac{2}{5}$.$\left(\frac{8g-14f}{25}ight)^2 + \left(\frac{12f-14g}{25}ight)^2 = \frac{2}{5}$$(64g^2 + 196f^2 - 224fg + 144f^2 + 196g^2 - 336fg) = \frac{2}{5} 	imes 625 = 250$$260g^2 + 340f^2 - 560fg = 250$$26g^2 + 34f^2 - 56fg = 25$  ---(iv)From $(i)$,$6f^2 + 4g^2 - 14fg = -25$,so $24f^2 + 16g^2 - 56fg = -100$  ---$(v)$Subtracting $(v)$ from (iv):$(26g^2 - 16g^2) + (34f^2 - 24f^2) = 25 - (-100)$$10g^2 + 10f^2 = 125$$f^2 + g^2 = \frac{125}{10} = \frac{25}{2}$

If the equation $3x^2+7xy+2y^2+2gx+2fy+2=0$ represents a pair of intersecting lines and the square of the distance of their point of intersection from the origin is $\frac{2}{5}$,then $f^2+g^2=$

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