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.