Question:

For a hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] if the length of the transverse axis is \(8\) and the distance between the foci is \(2\sqrt{41}\), then the length of its latus rectum is:

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For the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), remember: \[ c^2=a^2+b^2 \] and length of latus rectum is \[ \frac{2b^2}{a}. \]
Updated On: Jun 26, 2026
  • \(\dfrac{25}{2}\)
  • \(\dfrac{32}{5}\)
  • \(\dfrac{25}{4}\)
  • \(\dfrac{16}{5}\)
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The Correct Option is A

Solution and Explanation

Step 1: Use the transverse axis length.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] length of transverse axis is \[ 2a. \] Given, \[ 2a=8. \] Hence, \[ a=4. \]

Step 2: Use the distance between foci.
Distance between foci is \[ 2c. \] Given, \[ 2c=2\sqrt{41}. \] Therefore, \[ c=\sqrt{41}. \]

Step 3: Find \(b^2\).
For a hyperbola, \[ c^2=a^2+b^2. \] So, \[ 41=16+b^2. \] Hence, \[ b^2=25. \]

Step 4: Find the length of latus rectum.
Length of latus rectum of hyperbola is \[ \frac{2b^2}{a}. \] Therefore, \[ \frac{2b^2}{a} = \frac{2(25)}{4} = \frac{50}{4} = \frac{25}{2}. \]

Step 5: Final conclusion.
Hence, the length of latus rectum is \[ \boxed{\frac{25}{2}}. \]
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