1 \(2 x^2+5 x y+2 y^2+4 x+5 y+2=0\)
2 \(2 x^2+5 x y+2 y^2+4 x+5 y-2=0\)
3 \(2 x^2+5 x y+2 y^2=0\)
4 None of these
Explanation:
A We have equation of hyperbola \(2 x^2+5 x y+2 y^2+4 x+5 y+2=0\)
Let the equation of asymptotes be
\(2 \mathrm{x}^2+5 \mathrm{xy}+2 \mathrm{y}^2+4 \mathrm{x}+5 \mathrm{y}+\lambda=0\)
Here, \(\mathrm{a}=2, \mathrm{~b}=2, \mathrm{~h}=\frac{5}{2}, \mathrm{~g}=2, \mathrm{f}=\frac{5}{2}, \mathrm{c}=\lambda\)
This equation represents a pair of straight lines,
\(\therefore \Delta=0 \Rightarrow \mathrm{abc}+2 \mathrm{fgh}-\mathrm{af}^2-\mathrm{bg}^2-\mathrm{ch}^2=0\)
\(\therefore 4 \lambda+25-\frac{25}{2}-8-\lambda \times \frac{25}{4}=0\)
\(\Rightarrow \quad 4 \lambda-\frac{25 \lambda}{4}=\frac{25}{2}+8-25\)
\(\Rightarrow \quad \frac{16 \lambda-25 \lambda}{4}=\frac{25+16-50}{2}\)
\(\Rightarrow \quad \frac{-9 \lambda}{2}=16-25\)
\(\Rightarrow \quad-9 \lambda=-9 \times 2\)
\(\lambda=2\)
Putting the value of \(\lambda\) in equation (i), we get-
\(2 x^2+5 x y+2 y^2+4 x+5 y+2=0\)