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Question: The distance between the foci of the hyperbola \({x^2} - 3{y^2} - 4x - 6y - 11 = 0\) is \(A)4\) ...

The distance between the foci of the hyperbola x2−3y2−4x−6y−11=0{x^2} - 3{y^2} - 4x - 6y - 11 = 0 is
A)4A)4
B)6B)6
C)8C)8
D)10D)10

Explanation

Solution

First, we will see the concept of hyperbola.
Which is the set of all points in the given 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 given hyperbola.
Also, the eccentricity of the hyperbola is greater than 11. Eccentricity is denoted by ee
Formula used:
The hyperbola e=1+b2a2e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} where a,ba,b are the lengths of the semi-minor axes in the hyperbola respectively.
The distance between the foci for the hyperbola is 2ae2ae.

Complete step by step answer:
Since from the given that we have the equation as x2−3y2−4x−6y−11=0{x^2} - 3{y^2} - 4x - 6y - 11 = 0
Let us separate the variables of x and y separately, and also the constant values at the right-hand side, then we get x2−3y2−4x−6y−11=0⇒(x2−4x)−(3y2+6y)=11{x^2} - 3{y^2} - 4x - 6y - 11 = 0 \Rightarrow ({x^2} - 4x) - (3{y^2} + 6y) = 11
To get a generalized form of the algebraic expansion, add the number 44and also subtract the number 33on both sides of the equation we get (x2−4x)−(3y2+6y)=11⇒(x2−4x+4)−(3y2+6y+3)=11+4−3({x^2} - 4x) - (3{y^2} + 6y) = 11 \Rightarrow ({x^2} - 4x + 4) - (3{y^2} + 6y + 3) = 11 + 4 - 3 (since minus three directly applied into the brackets)
Further solving we get (x2−4x+4)−3(y2+y+1)=12({x^2} - 4x + 4) - 3({y^2} + y + 1) = 12 (taking common values out)
Since we know the algebraic formulas (a+b)2=a2+b2+2ab,(a−b)2=a2+b2−2ab{(a + b)^2} = {a^2} + {b^2} + 2ab,{(a - b)^2} = {a^2} + {b^2} - 2ab
Applying these values, we get (x2−4x+4)−3(y2+y+1)=12⇒(x−2)2−3(y+1)2=12({x^2} - 4x + 4) - 3({y^2} + y + 1) = 12 \Rightarrow {(x - 2)^2} - 3{(y + 1)^2} = 12
Now divide both sides using the number 1212then we get (x−2)2−3(y+1)2=12⇒(x−2)212−(y+1)24=1{(x - 2)^2} - 3{(y + 1)^2} = 12 \Rightarrow \dfrac{{{{(x - 2)}^2}}}{{12}} - \dfrac{{{{(y + 1)}^2}}}{4} = 1
Since we know that the equations of the hyperbola opening left and right in standard form is (x−h)2a2−(y−k)2b2=1\dfrac{{{{(x - h)}^2}}}{{{a^2}}} - \dfrac{{{{(y - k)}^2}}}{{{b^2}}} = 1
Comparing this we get the values of the a,ba,b. Thus, we have a2=12,b2=4{a^2} = 12,{b^2} = 4
Since the eccentricity is given as e=1+b2a2e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} and applying the values we get e=1+b2a2⇒1+412e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \Rightarrow \sqrt {1 + \dfrac{4}{{12}}}
Further solving we get e=1+b2a2⇒1612⇒423⇒23e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \Rightarrow \sqrt {\dfrac{{16}}{{12}}} \Rightarrow \dfrac{4}{{2\sqrt 3 }} \Rightarrow \dfrac{2}{{\sqrt 3 }}
Hence the distance between the foci is given as 2ae2ae.
Thus, we get 2ae=2×12×23⇒2×2×2⇒82ae = 2 \times \sqrt {12} \times \dfrac{2}{{\sqrt 3 }} \Rightarrow 2 \times 2 \times 2 \Rightarrow 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 the variables a,ba,b easily. After finding the values of a,ba,b we can find the eccentricity also.
After finding the values of the eccentricity and using the values of a, we can find the distance between the foci and which is the required answer.