Conic Sections — MATHS STD 11 Science — Question

2. ${\frac{\sqrt5}{2}}$

Solution:

The equation of the hyperbola is x² - 4y² = 1.

This can be rewritten in the following way:

$\frac{\text{x}^2}{1}-\frac{\text{y}^2}{\frac{1}{4}}=1$

This is the standard form of a hyperbola, where a = 1 and $\text{b}^2=\frac{1}{4}.$

The value of eccentricity is calculated in the following way:

$\text{b}^2=\text{a}^2(\text{e}^2-1)$

$\Rightarrow\frac{1}{4}=(\text{e}^2-1)$

$\Rightarrow\text{e}^2=\frac{5}{4}$

$\Rightarrow\text{e}=\frac{\sqrt5}{4}$