Question

If the eccentricity $e$ of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, passing through $(6, 4\sqrt{3})$, satisfies $15(e^2 + 1) = 34e$, then the length of the latus rectum of the hyperbola $\frac{x^2}{b^2} - \frac{y^2}{2(a^2 + 1)} = 1$ is:

• A. 10
• B. 20
• C. 25
• D. 30

Hint:

Solve the quadratic for $e>1$, find the ratio of $b^2/a^2$, substitute the point to find the values of $a$ and $b$, then apply the formula for the length of the latus rectum.

Answer 1

The Correct Option is A

We begin by finding $e$ from $15e^2 - 34e + 15 = 0$. Factoring the quadratic:
$$ (3e - 5)(5e - 3) = 0 $$
Since $e>1$, we take $e = 5/3$.
The relation between $a, b, e$ is $b^2 = a^2(e^2 - 1)$.
$$ b^2 = a^2 \left( \frac{25}{9} - 1 \right) = \frac{16}{9} a^2 $$
Substituting the point $(6, 4\sqrt{3})$ into the hyperbola equation:
$$ \frac{36}{a^2} - \frac{48}{b^2} = 1 $$
Substitute $b^2 = \frac{16}{9} a^2$:
$$ \frac{36}{a^2} - \frac{48}{\frac{16}{9}a^2} = 1 \implies \frac{36 - 27}{a^2} = 1 \implies a^2 = 9 $$
Consequently, $b^2 = \frac{16}{9} \cdot 9 = 16$.
For the new hyperbola $\frac{x^2}{b^2} - \frac{y^2}{2(a^2 + 1)} = 1$, we have $A^2 = b^2 = 16$ and $B^2 = 2(a^2+1) = 2(10) = 20$.
The length of the latus rectum for a standard hyperbola $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ is defined as $LR = \frac{2B^2}{A}$.
$$ LR = \frac{2 \times 20}{\sqrt{16}} = \frac{40}{4} = 10 $$