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If the equation \mathrm{a x^2+2 h x y+b y^2+2 g x+2 f y+c=0}  represents a pair of parallel lines, then

 

Option: 1

\mathrm{a=h=b}


Option: 2

\mathrm{g=f=c}


Option: 3

\mathrm{\frac{a}{h}=\frac{h}{b}=\frac{g}{f}}


Option: 4

none of these


Answers (1)

best_answer

Let \mathrm{Ix+my+n=0} and \mathrm{Ix+my+n_{1}=0} 

 represent parallel lines given by 

\mathrm{ a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 }\mathrm{ a x^2+2 h x y+b y^2+2 g x+2 f y+c=(l x+m y+n)\left(l x+m y+n_1\right)}

\therefore \mathrm{\quad l^2=a }--------------(i)

        \mathrm{ m^{2}=b} ------------(ii)

        \mathrm{nn_{1}=c} ----------(iii)

\mathrm{m(n+n_{1})=2f} ----------(iv)

    \mathrm{l(n+n_{1})=2g} ---------(v)

               \mathrm{2lm=2h} ---------(vi)

From (vi) \mathrm{\Rightarrow h=lm}\; \; \therefore \mathrm{h^{2}=l^{2}m^{2}=ab}

\therefore \mathrm{\frac{a}{h}=\frac{h}{b}} --------------(vii)

Also \mathrm{\frac{2 g}{2 f}=\frac{l\left(n+n_1\right)}{m\left(n+n_1\right)} \Rightarrow \frac{g}{f}=\frac{l}{m}=\frac{l m}{m^2}=\frac{h}{b}}

\therefore \mathrm{\frac{h}{b}=\frac{g}{f}}

From (vii) and (viii) \mathrm{\Rightarrow \quad \frac{a}{h}=\frac{h}{b}=\frac{g}{f}}

 

 

 

 

Posted by

Divya Prakash Singh

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