Question

If a hyperbola has asymptotes \(3x-4y-1=0\) and \(4x-3y-6=0\), then the transverse and conjugate axes of that hyperbola are

• A. \(x+y-5=0, x-y-1=0\)
• B. \(4x-3y=0, 3x+4y=0\)
• C. \(3x-4y=0, 4x+3y=0\)
• D. \(x+2y-1=0, 2x-y+1=0\)

Hint:

The transverse and conjugate axes of a hyperbola are the angle bisectors of its asymptotes. If the asymptotes are given by \(L_1=0\) and \(L_2=0\), their angle bisectors are given by \(\frac{L_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{L_2}{\sqrt{a_2^2+b_2^2}}\). These two equations represent the transverse and conjugate axes.

Answer 1

The Correct Option is A

Step 1: Understand the properties of asymptotes and axes of a hyperbola.
For a hyperbola, the transverse axis and conjugate axis are the bisectors of the angle between the asymptotes.
Let the equations of the asymptotes be \(L_1 = 3x - 4y - 1 = 0\) and \(L_2 = 4x - 3y - 6 = 0\).
The equations of the angle bisectors of two lines \(L_1 = a_1x + b_1y + c_1 = 0\) and \(L_2 = a_2x + b_2y + c_2 = 0\) are given by:
\[ \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}} \] Step 2: Calculate the denominators for the bisector equations.
For \(L_1 = 3x - 4y - 1 = 0\):
\(\sqrt{a_1^2 + b_1^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\).
For \(L_2 = 4x - 3y - 6 = 0\):
\(\sqrt{a_2^2 + b_2^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\). 
Step 3: Find the equations of the angle bisectors.
Substitute the values into the bisector formula: \[ \frac{3x - 4y - 1}{5} = \pm \frac{4x - 3y - 6}{5} \] Multiply by 5: \[ 3x - 4y - 1 = \pm (4x - 3y - 6) \] Case 1: Using the '+' sign}
\(3x - 4y - 1 = 4x - 3y - 6\)
Rearrange terms:
\(0 = 4x - 3x - 3y + 4y - 6 + 1\)
\(0 = x + y - 5\)
So, one axis is \(x + y - 5 = 0\).
Case 2: Using the '-' sign}
\(3x - 4y - 1 = -(4x - 3y - 6)\)
\(3x - 4y - 1 = -4x + 3y + 6\)
Rearrange terms:
\(3x + 4x - 4y - 3y - 1 - 6 = 0\)
\(7x - 7y - 7 = 0\)
Divide by 7:
\(x - y - 1 = 0\).
So, the equations of the angle bisectors are \(x + y - 5 = 0\) and \(x - y - 1 = 0\). These two lines represent the transverse and conjugate axes of the hyperbola. 
Step 4: Verify with the options.
The derived equations are \(x+y-5=0\) and \(x-y-1=0\). This exactly matches Option (1).