Question

The locus of the point of intersection of the lines \[\sqrt 3 x - y - 4\sqrt 3 t = 0\] & \[\sqrt 3 tx + ty - 4\sqrt 3 = 0\;\] (where t is a parameter) is a hyperbola. whose eccentricity is
A. \[\sqrt 3 \]
B. \[2\]
C. \[\dfrac{2}{{\sqrt 3 }}\]
D. \[\dfrac{4}{3}\]

Answer 1

Hint: To find the eccentricity of the hyperbola first we need to find the equation of hyperbola. The equation of the hyperbola can be found by equating the value of t in both the equations of the line given.

Complete step-by-step answer:
Given the equation of first line
​ \[\sqrt 3 x - y - 4\sqrt 3 t = 0\]
 Now find the value of t in terms of x and y
 \[\sqrt 3 x - y = 4\sqrt 3 t\]
Or,
 \[ \Rightarrow t = \dfrac{{\sqrt {3} x - y}}{{4\sqrt 3 }}\;\] ............. (1)
Now work on the second line and find the value of t
 \[\sqrt 3 tx + ty - 4\sqrt 3 = 0\;\]
Taking t common from the above we get
 \[t\left( {\sqrt {3} x + y} \right) = 4\sqrt 3 \]
or,
 \[ \Rightarrow t = \dfrac{{4\sqrt {3} }}{{\sqrt 3 x + y}}\;\] ............. (2)
From (1) and (2), The locus of the point of intersection of the lines is,
 \[\dfrac{{\sqrt 3 x - y}}{{4\sqrt 3 }} = \dfrac{{4\sqrt 3 }}{{\sqrt {3} x + y}}\]
On cross multiplying we get,
 \[3{x^2} - {y^2} = 48\]
Or,
 \[\Rightarrow \dfrac{{{x^2}}}{{16}} - \dfrac{{{y^{2}}}}{{48}} = 1\]
The above equation is a hyperbola of the form \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\]
Here the value of \[{a^2}\] and \[{b^2}\] is 16 and 48 respectively
We know the eccentricity of the hyperbola is
Eccentricity, \[e = \dfrac{{\sqrt {{a^2} + {b^2}} }}{a}\] ​​
On putting the value of \[{a^2}\] and \[{b^2}\]
We get the value of the eccentricity of hyperbola
 \[e = \dfrac{{\sqrt {16 + 48} }}{4} = \dfrac{{\sqrt {64} }}{4} = 2\]
So, the correct answer is “2”.

Note: The eccentricity of hyperbola is always greater than 1, the eccentricity of the ellipse is less than 1 while the eccentricity of the parabola is 1. That is one of the main differences between these conic figures.