Question

The line segment joining the points $(-3,\,1)$ and $(1,\,1)$ is the transverse axis of a hyperbola. If the length of the conjugate axis is 4, then the equation of the hyperbola is

• A. $(x+2)^2 - (y-1)^2 = 4$
• B. $(x+1)^2 - (y-1)^2 = 16$
• C. $(x+1)^2 - (y-1)^2 = 4$
• D. $(x+2)^2 - (y-1)^2 = 16$
• E. $(x-1)^2 - (y+1)^2 = 4$

Hint:

The transverse axis of a hyperbola connects the two vertices. Its midpoint is the centre. Always identify whether the transverse axis is horizontal or vertical before writing the standard form.

Answer 1

The Correct Option is C

Step 1: Concept:
• The centre of a hyperbola is the midpoint of the transverse axis.
• Length of transverse axis = \(2a\), and conjugate axis = \(2b\).

Step 2: Find Centre:
• Given points: \((-3,\,1)\) and \((1,\,1)\)
• Midpoint: \[ \left(\frac{-3+1}{2},\, \frac{1+1}{2}\right) = (-1,\,1) \]
• So, centre = \((-1,\,1)\)

Step 3: Find \(a\) and \(b\):
• Length of transverse axis: \[ |1 - (-3)| = 4 \Rightarrow 2a = 4 \Rightarrow a = 2 \Rightarrow a^2 = 4 \]
• Length of conjugate axis: \[ 2b = 4 \Rightarrow b = 2 \Rightarrow b^2 = 4 \]

Step 4: Form Equation:
• Standard form (horizontal hyperbola): \[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]
• Substituting \((h,k)=(-1,1)\), \(a^2=4\), \(b^2=4\): \[ \frac{(x+1)^2}{4} - \frac{(y-1)^2}{4} = 1 \]
• Simplified form: \[ (x+1)^2 - (y-1)^2 = 4 \]

Step 5: Final Answer:
• \[ (x+1)^2 - (y-1)^2 = 4 \]