Question:
The distance between the foci of the hyperbola $x^2 - 3y^2 - 4x - 6y -11 = 0$ is
The distance between the foci of the hyperbola $x^2 - 3y^2 - 4x - 6y -11 = 0$ is
Updated On: Jun 18, 2022
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The Correct Option is C
Solution and Explanation
$x^{2}-3 y^{2}-4 x-6 y-11=0$
$\Rightarrow\left(x^{2}-4 x\right)-3\left(y^{2}+2 y\right)=11$
$\Rightarrow\left(x^{2}-4 x+4\right)-3\left(y^{2}+2 y+1\right)=11+1$
$\Rightarrow \frac{(x-2)^{2}}{12}-\frac{(y+1)^{2}}{4}=1$
$\Rightarrow a^{2}=12, b^{2}=4$
$\therefore e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\frac{2}{\sqrt{3}}$
Therefore distance between focii is
$=2 ae =2 \times 2 \sqrt{3} \times \frac{2}{\sqrt{3}}$
$=8$
$\Rightarrow\left(x^{2}-4 x\right)-3\left(y^{2}+2 y\right)=11$
$\Rightarrow\left(x^{2}-4 x+4\right)-3\left(y^{2}+2 y+1\right)=11+1$
$\Rightarrow \frac{(x-2)^{2}}{12}-\frac{(y+1)^{2}}{4}=1$
$\Rightarrow a^{2}=12, b^{2}=4$
$\therefore e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\frac{2}{\sqrt{3}}$
Therefore distance between focii is
$=2 ae =2 \times 2 \sqrt{3} \times \frac{2}{\sqrt{3}}$
$=8$
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Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
