Question

The locus of the point of intersection of the lines $\sqrt{3}x-y-4\sqrt{3}t=0 {\text{ and}} \sqrt{3}tx+ty-4\sqrt{3}=0$ (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:From the two lines given in the question, take one of the equation of a line and write t in terms of x and y then substitute this value of t in the other equation and then rearrange the expressions in the form of the equation of a hyperbola and then find the eccentricity of the hyperbola.

Complete step-by-step answer:
In the above problem, two equations are given as:
$\begin{aligned} & \sqrt{3}x-y-4\sqrt{3}t=0.........\text{Eq}\text{.(1)} \\\ & \sqrt{3}tx+ty-4\sqrt{3}=0.........\text{Eq}\text{.(2)} \\\ \end{aligned}$
Now, writing the value of t in terms of x and y in eq. (2) as follows:
$\begin{aligned} & \sqrt{3}tx+ty-4\sqrt{3}=0 \\\ & \Rightarrow t\left( \sqrt{3}x+y \right)=4\sqrt{3} \\\ & \Rightarrow t=\dfrac{4\sqrt{3}}{\sqrt{3}x+y} \\\ \end{aligned}$
Substituting the above value of t in eq. (1) we get,
$\begin{aligned} & \sqrt{3}x-y=4\sqrt{3}\left( \dfrac{4\sqrt{3}}{\sqrt{3}x+y} \right) \\\ & \Rightarrow \left( \sqrt{3}x-y \right)\left( \sqrt{3}x+y \right)=48 \\\ \end{aligned}$
In the above equation, we will use the identity $(a – b)(a + b) = a^2 – b^2$.
$3{{x}^{2}}-{{y}^{2}}=48$
Now, rewriting the above equation in the form of a hyperbola$\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$because it is given that locus of the point of intersection of two lines is a hyperbola.
Dividing the above expression by 48 on both the sides we get,
$\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{48}=1$
We know that, the eccentricity of a hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is ${{e}^{2}}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}}$.
Comparing the two hyperbolas:
$\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{48}=1$
$\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
From the above two equations, we get:
$a^2 = 16$ and $b^2 = 48$
Substituting the values of $a^2$ and $b^2$ in the formula of eccentricity we get,
${{e}^{2}}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}}$
$\begin{aligned} & {{e}^{2}}=1+\dfrac{48}{16} \\\ & \Rightarrow {{e}^{2}}=\dfrac{16+48}{16} \\\ & \Rightarrow {{e}^{2}}=\dfrac{64}{16} \\\ & \Rightarrow {{e}^{2}}=4 \\\ & \Rightarrow e=\pm 2 \\\ \end{aligned}$
So, the eccentricity of hyperbola is $+2$ because eccentricity of the hyperbola is always greater than 1.
Hence, the correct option is (b).

Note: You might be thinking how the eccentricity of a hyperbola is greater than 1.
The formula of eccentricity is$e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$. The ratio of $b^2$ and $a^2$ can never be negative because squares of any number cannot be negative. So, the expression in the square root is always greater than 1. And the square root of a number greater than 1 is always greater than 1.