Question

If \(e_1\) and \(e_2\) are the eccentricities of the hyperbola \[ 16x^2-9y^2=1 \] and its conjugate respectively, then \(3e_1=\)

• A. \(5e_2\)
• B. \(4e_2\)
• C. \(2e_2\)
• D. \(e_2\)

Hint:

For a hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the eccentricity is \[ e=\sqrt{1+\frac{b^2}{a^2}} \] Always rewrite the equation in standard form before applying formulas.

Answer 1

The Correct Option is B

Step 1: Write the hyperbola in standard form.
$16x^2-9y^2=1 \implies x^2/(1/16)-y^2/(1/9)=1$. So $a^2=1/16$, $b^2=1/9$.
Step 2: Compute $e_1$.
$e_1=\sqrt{1+b^2/a^2}=\sqrt{1+16/9}=\sqrt{25/9}=5/3$.
Step 3: Identify the conjugate hyperbola.
Conjugate: $y^2/(1/9)-x^2/(1/16)=1$, i.e., $9y^2-16x^2=1$. Here $A^2=1/9$, $B^2=1/16$.
Step 4: Compute $e_2$.
$e_2=\sqrt{1+B^2/A^2}=\sqrt{1+9/16}=\sqrt{25/16}=5/4$.
Step 5: Check the relation.
$3e_1=3\times 5/3=5$ and $4e_2=4\times 5/4=5$. So $3e_1=4e_2$.
Step 6: State the answer.
\[ \boxed{3e_1=4e_2} \]