Question

The distance between the foci of the hyperbola $x^2-3y^2-4x-6y-11=0$ is

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is C

$8$

Calculating the distance between the foci of the hyperbola:

Given, the hyperbola is $x^2-3y^2-4x-6y-11=0$

$$\begin{gathered} \Rightarrow x^2-3y-4x-6y-11=0 \\ \Rightarrow \left(x^2-4x\right)-3\left(y^2+2y\right)=11 \\ \Rightarrow \left(x^2-4x+4-4\right)-3\left(y^2+2y+1-1\right)=11 \\ \Rightarrow \left[{\left(x-2\right)}^2-4\right]-3\left({\left(y+1\right)}^2-1\right)=11 \\ \Rightarrow {\left(x-2\right)}^2-4-3{\left(y+1\right)}^2+3=11 \\ \Rightarrow {\left(x-2\right)}^2-3{\left(y+1\right)}^2=12 \\ \Rightarrow \frac{{\left(x-2\right)}^2}{12}-\frac{{\left(y+1\right)}^2}{4}=1 \end{gathered}$$

The general form of the hyperbola is,

$$\begin{gathered} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \therefore \mathrm{a}=\sqrt{12},\mathrm{b}=2 \end{gathered}$$

On comparing with $\left[\frac{{\left(x-2\right)}^2}{12}-\frac{{\left(y+1\right)}^2}{4}=1\right]$

Calculating the eccentricity,

$$\begin{gathered} e=\sqrt{1+\frac{b^2}{a^2}} \\ =\sqrt{1+\frac{4}{12}} \\ =\frac{2}{\sqrt{3}} \end{gathered}$$

The distance between the foci is given by,

$$\begin{gathered} =2ae \\ =2\times 2\sqrt{3}\times \frac{2}{\sqrt{3}} \\ =8 \end{gathered}$$

Hence, the correct answer is option (C).