Conic Sections Question 87
Question: The combined equation of the asymptotes of the hyperbola $ 2x^{2}+5xy+2y^{2}+4x+5y=0 $
[Karnataka CET 2002]
Options:
A) $ 2x^{2}+5xy+2y^{2}=0 $
B) $ 2x^{2}+5xy+2y^{2}-4x+5y+2=0 $
C) $ 2x^{2}+5xy+2y^{2}+4x+5y-2=0 $
D) $ 2x^{2}+5xy+2y^{2}+4x+5y+2=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
Given, equation of hyperbola $ 2x^{2}+5xy+2y^{2}+4x+5y=0 $ and equation of asymptotes $ 2x^{2}+5xy+2y^{2}+4x+5y+\lambda =0 $ …….(i), which is the equation of a pair of straight lines. We know that the standard equation of a pair of straight lines is $ ax^{2}+2hxy+by^{2}+2gx+2fy+c=0. $ Comparing equation (i) with standard equation, we get $ a=2,b=2, $
$ h=\frac{5}{2},g=2,f=\frac{5}{2} $ and $ c=\lambda . $
We also know that the condition for a pair of straight lines is $ abc+2fgh-af^{2}-bg^{2}-ch^{2}=0. $
Therefore $ 4\lambda +25-\frac{25}{2}-8-\frac{25}{4}\lambda =0 $
or $ -\frac{9\lambda }{4}+\frac{9}{2}=0 $ or $ \lambda =2 $ . Substituting value of $ \lambda $ in equation (i), we get $ 2x^{2}+5xy+2y^{2}+4x+5y+2=0. $
BETA
