If the equation $3 x^2+7 x y+2 y^2+2 g x+2 f y+2=0$ 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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Aicharya · Accepted Answer

The equation,

represents a pair of intersecting lines, so $ \begin{aligned} & \text { (3) }(2)(2)+7(g)(f)-3 f^2-2 g^2-2\left(\frac{7}{2}\right)^2=0 \\ & \Rightarrow \quad 24+14 g f-6 f^2-4 g^2-49=0 \end{aligned} $

To get the point of intersection of pair of straight lines, we have to differentiate partially the Eq. (i) with respect to $x$ and $y$ respectively. then we are getting

Now, point of intersection of eqs. (iii) and (iv) is the point of intersection of pair of straight line Eq. (i). $ \left(\frac{14 f-8 g}{25}, \frac{14 g-12 f}{25}\right) $ Now, according to the question, $ \begin{aligned} & \left(\frac{14 f-8 g}{25}\right)^2+\left(\frac{14 g-12 f}{25}\right)^2=\frac{2}{5} \\ \Rightarrow \quad 196 f^2+64 g^2-224 f g+196 g^2 & +144 f^2 \\ \Rightarrow \quad & \quad 260 g^2+340 f^2-560 g f=250 \end{aligned} $

From Eqs. (ii) and (v) $ \begin{gathered} 26 g^2+34 f^2-56 g f=25 \\ 16 g^2+24 f^2-56 g f=-100 \\ -\frac{-\quad+\quad+}{10 g^2+10 f^2=125}+ \\ g^2+f^2=\frac{25}{2} \end{gathered} $

If the equation $3 x^2+7 x y+2 y^2+2 g x+2 f y+2=0$ represents a pair of intersecting lines and the square…

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