Question

If the eccentricity of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is \( \frac{5}{4} \) and \( 2x + 3y - 6 = 0 \) is a focal chord of the hyperbola, then the length of transverse axis is equal to:

• A. \( \frac{12}{5} \)
• B. \( 6 \)
• C. \( \frac{24}{7} \)
• D. \( \frac{24}{5} \)
• E. \( \frac{12}{7} \)

Hint:

To find the focus from a focal chord equation, simply find where the line intersects the major/transverse axis (usually by setting $y=0$ or $x=0$).

Answer 1

The Correct Option is D

Concept: A focal chord is a line segment that passes through one of the foci of the conic. For the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the foci are located at \( (\pm ae, 0) \).

Step 1: Find the focus using the line equation.
The line \( 2x + 3y - 6 = 0 \) passes through a focus. Since the foci lie on the x-axis (\( y=0 \)), we find the x-intercept of the line: \[ 2x + 3(0) - 6 = 0 \] \[ 2x = 6 \quad \Rightarrow \quad x = 3 \] Thus, the focus is at \( (3, 0) \). This means \( ae = 3 \).

Step 2: Solve for \( a \).
We are given eccentricity \( e = \frac{5}{4} \). \[ a\left(\frac{5}{4}\right) = 3 \] \[ a = 3 \times \frac{4}{5} = \frac{12}{5} \]

Step 3: Calculate the length of the transverse axis.
Length of transverse axis = \( 2a \) \[ 2a = 2 \times \frac{12}{5} = \frac{24}{5} \]