Q.The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is
(A) $\dfrac{4}{3}$
(B) $\dfrac{4}{\sqrt{3}}$
(C) $\dfrac{2}{\sqrt{3}}$
(D) none of these

The key idea is understanding the standard properties of a hyperbola.

Given the latus rectum $LR = 8$, we use the formula $LR = \frac{2b^2}{a}$, which gives $8 = \frac{2b^2}{a} \implies b^2 = 4a$.

The conjugate axis $2b$ is half the distance between the foci $2ae$. So, $2b = \frac{1}{2}(2ae) \implies 2b = ae$. From this, we can express $a$ as $a = \frac{2b}{e}$.

Substitute $a = \frac{2b}{e}$ into $b^2 = 4a$:

$b^2 = 4 \left(\frac{2b}{e}\right) \implies b^2 = \frac{8b}{e}$.

Since $b \neq 0$ for a hyperbola, we can divide by $b$ to get $b = \frac{8}{e}$.

Now, use the fundamental relation for a hyperbola, $b^2 = a^2(e^2 - 1)$. Substitute $b = \frac{8}{e}$ and $a = \frac{2b}{e} = \frac{2(8/e)}{e} = \frac{16}{e^2}$:

$\left(\frac{8}{e}\right)^2 = \left(\frac{16}{e^2}\right)^2 (e^2 - 1)$ …

To find the eccentricity of a hyperbola, we need to understand its fundamental properties and how they relate to the given information. The eccentricity, denoted by $e$, is a crucial parameter that defines the shape of a conic section. For a hyperbola, $e > 1$. The problem provides two pieces of information: the length of the latus rectum and a relationship between the conjugate axis and the distance between the foci. We will translate these geometric properties into algebraic equations involving the semi-major axis ($a$), semi-minor axis ($b$), and eccentricity ($e$).

Let's break down the problem step-by-step.

1. Recall the standard properties of a hyperbola.

   For a standard hyperbola with equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

   • The length of the latus rectum is $\frac{2b^2}{a}$.
   • The length of the conjugate axis is $2b$.
   • The distance between the foci is $2ae$.
   • The fundamental relationship between $a$, $b$, and $e$ is $b^2 = a^2(e^2 - 1)$. This equation is derived from the definition of eccentricity and the geometry of the hyperbola.

   For a hyperbola, the relationship between $a$, $b$, and $e$ is:

   $b^2 = a^2(e^2 - 1)$

2. Use the given information about the latus rectum.

   The problem states that the latus rectum is 8. Using the formula from Step 1:

   $\frac{2b^2}{a} = 8$

   This simplifies to:

   $b^2 = 4a \quad (*)$

   This gives us our first equation relating $a$ and $b$.

3. Use the given information about the conjugate axis and distance between foci.

   The problem states that the conjugate axis is equal to half of the distance between the foci.

   Length of conjugate axis $= 2b$

   Distance between foci $= 2ae$

   So, the condition is:

   $2b = \frac{1}{2}(2ae)$

   $2b = ae \quad (**)$

   This gives us our second equation, relating $a$, $b$, and $e$.

4. Solve the system of equations to find the eccentricity.

   We now have three key equations:

   (1) $b^2 = 4a$ (from latus rectum)

   (2) $2b = ae$ (from conjugate axis and foci distance)

   (3) $b^2 = a^2(e^2 - 1)$ (fundamental relation)

   Our goal is to find $e$. We can substitute (1) into (3) to eliminate $b^2$:

   $4a = a^2(e^2 - 1)$

   Since $a$ cannot be zero for a hyperbola, we can divide both sides by $a$:

   $4 = a(e^2 - 1) \quad (***)$

   Now, let's use equation (2). Square both sides of $2b = ae$:

   $(2b)^2 = (ae)^2$ …