Question

If the eccentricity and length of the latus rectum of a hyperbola are \( \frac{\sqrt{13}}{3} \) and \( \frac{10}{3} \) units respectively, then what is the length of the transverse axis?

• A. \( \frac{7}{2} \) unit
• B. \( 12 \) unit
• C. \( \frac{15}{2} \) unit
• D. \( \frac{15}{4} \) unit

Hint:

For hyperbolas, remember: - \( c^2 = a^2 + b^2 \) for standard forms. - The latus rectum formula \( \frac{2b^2}{a} \) helps find unknowns efficiently.

Answer 1

The Correct Option is C

Step 1: Formula for Latus Rectum For a hyperbola, the length of the latus rectum is given by: \[ \frac{2b^2}{a}. \] We are given: \[ \frac{2b^2}{a} = \frac{10}{3}. \] Step 2: Using the Eccentricity Formula The eccentricity of a hyperbola is given by: \[ e = \frac{c}{a}. \] We are given: \[ e = \frac{\sqrt{13}}{3}. \] From the standard hyperbola relation: \[ c^2 = a^2 + b^2. \] Step 3: Expressing \( c \) and \( b^2 \) From \( e = \frac{c}{a} \), we express \( c \) as: \[ c = \frac{\sqrt{13}}{3} a. \] Rearrange the standard relation: \[ \left(\frac{\sqrt{13}}{3} a\right)^2 = a^2 + b^2. \] Expanding: \[ \frac{13}{9} a^2 = a^2 + b^2. \] Rearrange: \[ b^2 = \frac{4}{9} a^2. \] Step 4: Solving for \( a \) Using the latus rectum equation: \[ \frac{2b^2}{a} = \frac{10}{3}. \] Substituting \( b^2 = \frac{4}{9} a^2 \): \[ \frac{2 \times \frac{4}{9} a^2}{a} = \frac{10}{3}. \] Simplify: \[ \frac{8}{9} a = \frac{10}{3}. \] Solving for \( a \): \[ a = \frac{10}{3} \times \frac{9}{8} = \frac{90}{24} = \frac{15}{4}. \] Step 5: Finding Transverse Axis Length The transverse axis length is: \[ 2a = 2 \times \frac{15}{4} = \frac{30}{4} = \frac{15}{2}. \] Thus, the correct answer is \( \frac{15}{2} \), which matches option (C).