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

First, we solve for the eccentricity $e$ from the given quadratic equation:
$$ 15e^2 - 34e + 15 = 0 $$
Using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$ e = \frac{34 \pm \sqrt{(-34)^2 - 4(15)(15)}}{2(15)} = \frac{34 \pm \sqrt{1156 - 900}}{30} = \frac{34 \pm \sqrt{256}}{30} = \frac{34 \pm 16}{30} $$
This gives $e = \frac{50}{30} = \frac{5}{3}$ or $e = \frac{18}{30} = \frac{3}{5}$.
Since for a hyperbola $e>1$, we must have $e = \frac{5}{3}$.
For a hyperbola, $e^2 = 1 + \frac{b^2}{a^2}$.
$$ \left(\frac{5}{3}\right)^2 = 1 + \frac{b^2}{a^2} \implies \frac{25}{9} = 1 + \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = \frac{16}{9} \implies b^2 = \frac{16a^2}{9} $$
The hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ passes through $(6, 4\sqrt{3})$, so:
$$ \frac{36}{a^2} - \frac{(4\sqrt{3})^2}{b^2} = 1 \implies \frac{36}{a^2} - \frac{48}{b^2} = 1 $$
Substitute $b^2 = \frac{16a^2}{9}$:
$$ \frac{36}{a^2} - \frac{48 \cdot 9}{16a^2} = 1 \implies \frac{36}{a^2} - \frac{27}{a^2} = 1 \implies \frac{9}{a^2} = 1 \implies a^2 = 9 $$
Then $b^2 = \frac{16 \cdot 9}{9} = 16$.
The second hyperbola equation is $\frac{x^2}{16} - \frac{y^2}{2(9 + 1)} = 1$, which simplifies to $\frac{x^2}{16} - \frac{y^2}{20} = 1$.
Here, $A^2 = 16$ (so $A = 4$) and $B^2 = 20$.
The length of the latus rectum is $\frac{2B^2}{A} = \frac{2(20)}{4} = \frac{40}{4} = 10$.