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The asymptotes of the curve \mathrm{2 x^2+5 x y+2 y^2+4 x+5 y=0}are given by

Option: 1

\mathrm{ 2 x^2+5 x y+2 y^2+4 x+5 y+2=0}


Option: 2

\mathrm{2 x^2+5 x y+2 y^2+4 x+5 y-2=0}


Option: 3

\mathrm{2 x^2+5 x y+2 y^2+4 x+5 y-4=0}


Option: 4

None of these


Answers (1)

best_answer

The hyperbola is given by 

2 x^2+5 x y+2 y^2+4 x+5 y=0 ---------------(i)

Since the equation of hyperbola will differ from equation of asymptote by a constant. So, equation of asymptote is 

2 x^2+5 x y+2 y^2+4 x+5 y+\lambda=0 ---------------(ii)

If (ii) represents 2 straight lines, we must have

\mathrm{\begin{aligned} & a b c+2 f g h-a f^2-b g^2-c h^2=0 \\ & \text { or } 2.2 \lambda+2.5 / 2.4 / 2.5 / 2-2.25 / 4-2.4-\lambda(25 / 4)=0 \\ & \text { or } 9 \lambda=18 \therefore \lambda=2 \end{aligned}}

Putting value of \mathrm{\lambda}  in (ii) , equation of asymptotes is 

\mathrm{2 x^2+5 x y+2 y^2+4 x+5 y+2=0}

 

 

Posted by

Devendra Khairwa

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