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 asymptotes is given by \[S + \lambda = 0\left( {\lambda \in R} \right)\] which represent a pair of straight lines so \[\lambda \] can be determined. The equation of asymptotes is \[A = S + \lambda = 0\] if equation of conjugate hyperbola of the curve \[S = 0\] be represents by \[{S_1}\] , then A is arithmetic mean of the curves \[{S_1}\] and S.
​A hyperbola passing through origin has \[2x - y + 3 = 0\] and \[x - 2y + 2 = 0\] as its asymptotes, then equation of its transverse and conjugate axes are:
A \[x - y + 2 = 0\] and \[3x - 3y + 5 = 0\]
B \[x + y + 1 = 0\] and \[3x - 3y + 5 = 0\]
C \[x - y + 1 = 0\] and \[3x - 3y + 5 = 0\]
D \[2x - 2y + 1 = 0\] and \[3x - 3y + 5 = 0\]

Answer 1

Hint: Asymptotes of a hyperbola are the lines that pass through the center of the hyperbola. All hyperbolas have asymptotes, which are straight lines that form an X that the hyperbola approaches but never touches. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints, hence we can find the equation of its transverse and conjugate axes by applying the equation of bisector of angle of the asymptotes.

Complete step-by-step answer:
A hyperbola passing through origin has \[2x - y + 3 = 0\] and \[x - 2y + 2 = 0\] as its asymptotes.
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:
 \[\dfrac{{2x - y + 3}}{{\sqrt 5 }} = \pm \dfrac{{x - 2y + 2}}{{\sqrt 5 }}\]
 \[ \Rightarrow \] \[2x - y + 3 = \pm \left( {x - 2y + 2} \right)\]
 \[ \Rightarrow \] \[2x - y + 3 = x - 2y + 2\]
And
 \[2x - y + 3 = x - 2y + 2 = - \left( {x - 2y + 2} \right)\]
Simplifying the terms, we get
 \[ \Rightarrow \] \[x + y + 1 = 0\] and \[3x - 3y + 5 = 0\]
Hence, option C is the right answer.
So, the correct answer is “Option B”.

Note: We know that conjugate axes is the line through the center of an ellipse or a hyperbola and perpendicular to the line through the two foci, hence to find the equation of its transverse and conjugate axes we need to apply equation of bisector of angle of the asymptotes, the foci lie on the line that contains the transverse axis and the conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints.