Hyperbola Equation Standardization

The given equation is $6x^2 - 12y^2 = -36$. We need to convert this into a standard form to identify the hyperbola's parameters.

Divide the equation by $-36$:

$ \frac{6x^2}{-36} - \frac{12y^2}{-36} = \frac{-36}{-36} $

Simplify the terms:

$ -\frac{x^2}{6} + \frac{y^2}{3} = 1 $

Rearrange into the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$:

$ \frac{y^2}{3} - \frac{x^2}{6} = 1 $

Hyperbola Parameters Identification

By comparing $\frac{y^2}{3} - \frac{x^2}{6} = 1$ with the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, we identify:

• $a^2 = 3$, therefore $a = \sqrt{3}$
• $b^2 = 6$, therefore $b = \sqrt{6}$

Conjugate Axis Length Calculation

For a hyperbola oriented vertically ($\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$), the length of the conjugate axis is $2b$.

Length of conjugate axis = $2b = 2\sqrt{6}$.

Latus Rectum Length Calculation

The length of the latus rectum for any hyperbola is given by the formula $\frac{2b^2}{a}$.

Length of latus rectum = $ \frac{2b^2}{a} = \frac{2 \times 6}{\sqrt{3}} = \frac{12}{\sqrt{3}} $

To simplify, rationalize the denominator:

$ \frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3} $

Final Answer Determination

The calculated lengths are:

• Conjugate axis length: $2\sqrt{6}$
• Latus rectum length: $4\sqrt{3}$

These values match Option D.