Question

The hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) passes through the point \( (\sqrt{6}, 3) \) and the length of the latus rectum is \( \frac{18}{5} \). Then the length of the transverse axis is equal to:

• A. $5$
• B. $4$
• C. $3$
• D. $2$
• E. $1$

Hint:

In hyperbola problems, "transverse axis" always refers to the axis containing the foci (the positive term in the equation), and its length is $2a$. "Conjugate axis" is the other one, with length $2b$.

Answer 1

The Correct Option is D

Concept: For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:
• Length of latus rectum $= \frac{2b^2}{a}$.
• Length of transverse axis $= 2a$.
• Any point $(x, y)$ on the hyperbola must satisfy its equation.

Step 1: Use the latus rectum to express $b^2$ in terms of $a$.
\[ \frac{2b^2}{a} = \frac{18}{5} \quad \Rightarrow \quad b^2 = \frac{18a}{10} = \frac{9a}{5} \]

Step 2: Substitute the point and $b^2$ into the hyperbola equation.
The point is $(\sqrt{6}, 3)$, so $x^2 = 6$ and $y^2 = 9$: \[ \frac{6}{a^2} - \frac{9}{b^2} = 1 \] Substitute $b^2 = \frac{9a}{5}$: \[ \frac{6}{a^2} - \frac{9}{9a/5} = 1 \quad \Rightarrow \quad \frac{6}{a^2} - \frac{5}{a} = 1 \]

Step 3: Solve the equation for $a$.
Multiply by $a^2$: \[ 6 - 5a = a^2 \quad \Rightarrow \quad a^2 + 5a - 6 = 0 \] Factorizing: \[ (a + 6)(a - 1) = 0 \] Since $a$ must be positive, $a = 1$.

Step 4: Find the length of the transverse axis.
The length of the transverse axis is $2a$: \[ 2 \times 1 = 2 \]