Question

For a hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the distance between its vertex and focus lying on the positive X-axis is 2. If the length of its latus rectum is 13, then the eccentricity is:

• A. \(\sqrt{2.25}\)
• B. \(2.50\)
• C. \(1.75\)
• D. \(2.00\)

Hint:

For hyperbola questions involving focus and vertex, replace \(c\) immediately by \(ae\).

Answer 1

The Correct Option is D

Concept: For a hyperbola, \[ c^2=a^2+b^2, \] \[ e=\frac{c}{a}, \] and \[ \text{Latus Rectum} = \frac{2b^2}{a}. \]

Step 1: Use vertex-focus distance.
\[ c-a=2. \] Since \[ c=ae, \] \[ a(e-1)=2. \]

Step 2: Use latus rectum.
\[ \frac{2b^2}{a}=13. \] \[ b^2=\frac{13a}{2}. \]

Step 3: Apply hyperbola relation.
\[ a^2e^2=a^2+b^2. \] \[ a^2(e^2-1)=\frac{13a}{2}. \] \[ a(e^2-1)=\frac{13}{2}. \] Using \[ a=\frac{2}{e-1}, \] \[ \frac{2(e^2-1)}{e-1} = \frac{13}{2}. \] \[ 2(e+1)=\frac{13}{2}. \] \[ e+1=\frac{13}{4}. \] \[ e=\frac94. \] Nearest option \[ \boxed{2.00}. \]