  Question

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

A
2fgh=bg2+ch2  B
bg2ch2  C
abc=2fgh  D
None of these  Solution

The correct option is A $$2fgh= bg^{2}+ch^{2}$$As $$\displaystyle s=ax^{2}+2hxy +by^{2}+2gx +2fy+c=0$$ represent a pair of line $$\displaystyle \therefore \begin{vmatrix}a &h &g \\h &b &f \\g &f &c \end{vmatrix}=0$$or $$\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0....(1)$$ Now say point ofintersection onY axis be $$\displaystyle (0,y_{1}$$ and  point of intersection of pair of line be obtained by solving the equations $$\displaystyle \frac{\partial s}{\partial x}=0=\frac{\partial s}{\partial y}$$ $$\displaystyle \therefore \frac{\partial s}{\partial x}=0\Rightarrow ax+by+g=0$$ $$\displaystyle \begin{matrix}\Rightarrow \\ \Rightarrow \end{matrix} \left\{\begin{matrix}hy_{1}+g=0 \\by_{1}+f=0 \end{matrix}\right.> (*)$$ and $$\displaystyle \frac{\partial s}{\partial y}=0\Rightarrow bx+by+f=0$$  On compairing the equation given in (*) we get $$\displaystyle bg=fh$$ and  $$\displaystyle bg^{2}=fgh ....(2)$$ Again $$\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$$ meet at y-axis $$\displaystyle \therefore x=0$$ $$\displaystyle \Rightarrow by^{2}+2fy+c=0$$ whose roots must be equal $$\displaystyle \Rightarrow by^{2}+2fy+c=0$$ whose roots must be equal $$\displaystyle \therefore f^{2}=bc af^{2}=abc ......(3)$$ Now using (2) and (3) in equation (I) we have $$\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$$ $$\displaystyle \Rightarrow (abc-af^{2})+(fgh-bg^{2})+fgh-ch^{2}=0$$ $$\displaystyle \Rightarrow 0+0+fgh-ch^{2}=0 \therefore ch^{2}=fgh .....(4)$$ Now adding (2) and (4) $$\displaystyle 2fgh=ch^{2}+bg^{2}$$Maths

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