Question

For $6x^2 - 12y^2 = -36$ hyperbola, the lengths of conjugate axis and latus-rectum, respectively, are:

• A. $2\sqrt{3}$ and $4\sqrt{3}$
• B. $2\sqrt{6}$ and $\sqrt{6}$
• C. $2\sqrt{3}$ and $4\sqrt{6}$
• D. $2\sqrt{6}$ and $4\sqrt{3}$

Hint:

If the right side is $-1$ after standard division, it's a conjugate hyperbola. Be careful identifying $a^2$ and $b^2$ based on the sign, not magnitude.

Answer 1

The Correct Option is D

Concept: A hyperbola in standard form is either $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (horizontal) or $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ (vertical). For a vertical hyperbola, the conjugate axis length is $2a$ and the latus-rectum is $\frac{2a^2}{b}$.

Step 1: Convert to standard form.
Divide the equation $6x^2 - 12y^2 = -36$ by $-36$: \[ \frac{6x^2}{-36} - \frac{12y^2}{-36} = 1 \] \[ -\frac{x^2}{6} + \frac{y^2}{3} = 1 \quad \Rightarrow \quad \frac{y^2}{3} - \frac{x^2}{6} = 1 \] Here, $b^2 = 3$ and $a^2 = 6$. So, $b = \sqrt{3}$ and $a = \sqrt{6}$.

Step 2: Calculate the length of the conjugate axis.
For the vertical hyperbola $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$, the conjugate axis lies along the x-axis. Length of conjugate axis $= 2a = 2\sqrt{6}$.

Step 3: Calculate the length of the latus-rectum.
Length of latus-rectum $= \frac{2a^2}{b} = \frac{2(6)}{\sqrt{3}} = \frac{12}{\sqrt{3}}$. Rationalizing: $\frac{12\sqrt{3}}{3} = 4\sqrt{3}$.