© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Recall Standard Properties of a Hyperbola
Consider the standard hyperbola given by the equation
$ \frac{x^2}{a^2} \;-\; \frac{y^2}{b^2} \;=\; 1 $.
Its eccentricity is given by
$ e \;=\; \sqrt{1 \;+\; \frac{b^2}{a^2}}$.
Important related lengths are:
Latus Rectum: The length of the latus rectum is
$ \frac{2\,b^2}{a} $.
Conjugate Axis: The length of the conjugate axis is
$2\,b$.
Distance between Foci: The foci lie at $(\pm a e,\,0)$, making the distance between the foci equal to
$2\,a\,e$.
Step 2: Express the Given Conditions Mathematically
Latus Rectum Condition:
Given that the length of the latus rectum is $8$, we write:
$$
\frac{2\,b^2}{a} = 8.
$$
Conjugate Axis and Foci Condition:
Given that the length of the conjugate axis ($2\,b$) is half the distance between the foci ($2\,a\,e$), we have:
$$
2\,b = \frac{1}{2}\bigl(2\,a\,e\bigr)
\;\;\Longrightarrow\;\;
2\,b = a\,e
\;\;\Longrightarrow\;\;
b = \frac{a\,e}{2}.
$$
Step 3: Simplify the Latus Rectum Condition
From
$ \frac{2\,b^2}{a} = 8 $,
we get:
$$
2\,b^2 = 8\,a
\;\;\Longrightarrow\;\;
b^2 = 4\,a.
$$
Step 4: Use the Relationship $b^2 = a^2(e^2 - 1)$
For the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $, another well-known relationship is:
$$
b^2 = a^2 \bigl(e^2 - 1\bigr).
$$
We will connect this to the condition
$ b^2 = 4\,a $
once we find $a$ in terms of $e$, or $b$ in terms of $a$ and $e$.
Step 5: Express $b^2$ From $b = \frac{a\,e}{2}$
From the second condition:
$$
b = \frac{a\,e}{2}
\;\;\Longrightarrow\;\;
b^2 = \frac{a^2\,e^2}{4}.
$$
But we also know $b^2 = a^2 (e^2 - 1)$. Hence, equate the two expressions for $b^2$:
$$
a^2\bigl(e^2 - 1\bigr) = \frac{a^2\,e^2}{4}.
$$
Step 6: Solve for the Eccentricity $e$
Dividing both sides by $a^2$, we get:
$$
e^2 - 1 = \frac{e^2}{4}.
$$
Multiply through by 4:
$$
4\,e^2 - 4 = e^2
\;\;\Longrightarrow\;\;
3\,e^2 = 4
\;\;\Longrightarrow\;\;
e^2 = \frac{4}{3}
\;\;\Longrightarrow\;\;
e = \frac{2}{\sqrt{3}}.
$$
Step 7: Final Answer
The eccentricity of the given hyperbola is
$ \boxed{\frac{2}{\sqrt{3}}}. $
