Question

The distance between the foci of the hyperbola \[{x^2} - 3{y^2} - 4x - 6y - 11 = 0 \] is
A.4
B.6
C.8
D.10

Answer 1

Hint: A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distance between two fixed points on a plane stays constant. The two points are called the foci of the hyperbola. Also remember that the eccentricity of a hyperbola is greater than 1. Eccentricity is denoted by \[e \] . In hyperbola \[e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \] Where ‘a’ and ‘b’ are the lengths of semi major and semi minor axes in a hyperbola respectively.

Complete step-by-step answer:
Now given a polynomial, \[{x^2} - 3{y^2} - 4x - 6y - 11 = 0 \]
Separating x and y variable separately, shifting constant on right hand side we get,
 \[ \Rightarrow ({x^2} - 4x) - (3{y^2} + 6y + ) = 11 \]
Adding 4 on both side and subtract -3 on both side, we get,
 \[ \Rightarrow ({x^2} - 4x + 4) - (3{y^2} + 6y + ( - 3)) = 11 + 4 - 3 \]
 \[ \Rightarrow ({x^2} - 4x) - (3{y^2} + 6y + 3) = 11 + 1 \]
Taking 3 common on y variable group,
 \[ \Rightarrow ({x^2} - 4x + 4) - 3({y^2} + 2y + 1) = 12 \]
We can see that it is in the form \[{a^2} + {b^2} = {a^2} + {b^2} + 2ab \] and \[{a^2} - {b^2} = {a^2} + {b^2} - 2ab \] .
Applying these we get,
 \[ \Rightarrow \dfrac{{{{(x - 2)}^2}}}{{12}} - \dfrac{{3{{(y + 1)}^2}}}{{12}} = 1 \] .
Simple division,
 \[ \Rightarrow \dfrac{{{{(x - 2)}^2}}}{{12}} - \dfrac{{{{(y + 1)}^2}}}{4} = 1 \]
We know that the equation of a hyperbola opening left and right in standard form is:
 \[ \dfrac{{{{(x - h)}^2}}}{{{a^2}}} - \dfrac{{{{(y - k)}^2}}}{{{b^2}}} = 1 \] Where, the centre is \[(h,k) \] and the vertices are \[(h \pm a,k) \] .
Comparing this with the above equation we can find ‘a’ and ‘b’ values.
That is, \[ \Rightarrow {a^2} = 12,{b^2} = 4 \]
We know the eccentricity is given by
 \[e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \] Substituting
 \[ \Rightarrow e = \sqrt {1 + \dfrac{4}{{12}}} \]
 \[ \Rightarrow e = \sqrt { \dfrac{{16}}{{12}}} \]
 \[ \Rightarrow e = \dfrac{2}{{ \sqrt 3 }} \]
Therefore distance between foci is, \[2ae \] substituting we get,
 \[2ae = 2 \times \sqrt {12} \times \dfrac{2}{{ \sqrt 3 }} = 8 \]
So, the correct answer is “Option C”.

Note: All we do is convert the given equation into the standard hyperbola equation. So that we can find ‘a’ and ‘b’ values easily. Knowing the values of ‘a’ and ‘b’ we can find eccentricity. Now we know the values of eccentricity and ‘a’, using this we can find the distance between foci. Similarly we do it for any problem with different equations.