Question

# The second-degree curve and pair of asymptotes differ by a constant. Let the second-degree curve $$S = 0$$ represent the hyperbola then respective pair of asymptote is given by.$$\displaystyle S+\lambda =0\left ( \lambda \in R \right )$$ which represent a pair of straight lines so $$\lambda$$  can be determined. The equation of asymptotes is $$\displaystyle A=s+\lambda =0$$ if equation of conjugate hyperbola of the curve $$S =0$$ be represents by $$S_{1}$$, then $$A$$ is arithmetic mean of the curves $$S_{1}$$, & $$S$$.A hyperbola passing through origin has $$\displaystyle 2x-y+3=0$$ and $$\displaystyle x-2y+2=0$$ as its asymptotes, then equation of its transverse and conjugate axes are:

A
xy+2=0 and 3x3y+5=0
B
x+y+2=0 and 3x3y+5=0
C
xy+1=0 and 3x3y+5=0
D
2x2y+1=0 and 3x3y+5=0

Solution

## The correct option is C $$\displaystyle x-y+1=0$$ and $$\displaystyle 3x-3y+5=0$$The transverse axis of hyperbola is the bisector of the angle between the asymptotes containing the origin and the conjugate axis is the other bisector. And equation of bisector of angle of the asymptotes are given by$$\displaystyle \frac{2x-y+3}{\sqrt{5}}=\pm \frac{x-2y+2}{\sqrt{5}}$$$$\displaystyle \Rightarrow 2x-y+3 =\pm \left ( x-2y+2 \right )$$$$\displaystyle \Rightarrow 2x-y+3 =x-2y+2$$and $$\displaystyle 2x-y+3 =x-2y+2= -\left ( x-2y+2 \right )$$$$\displaystyle \Rightarrow x+y+1=0 \ and \ 3x-3y+5=0$$Hence, option 'C' is correct.Maths

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