Question

The locus of the point of intersection of the lines $\sqrt 3 x - y - 4\sqrt 3 t = 0$ 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: Here, it is given that the locus of point of intersection of the given two lines is a hyperbola. So firstly, find the standard equation of hyperbola by putting the value of $t$ in the second equation which is obtained from the first equation. Then by applying a standard relation between $a$, $b$ and $e$ for hyperbola that is ${b^2} = {a^2}\left( {{e^2} - 1} \right)$ where $a$ is the length of semi major axis, $b$ is the length of semi minor axis and $e$ is the eccentricity of the hyperbola, we can get the eccentricity of the hyperbola.

Complete step-by-step solution:
Given two lines are $\sqrt 3 x - y - 4\sqrt 3 t = 0$-----------(1)
and $\sqrt 3 tx + ty - 4\sqrt 3 = 0$-------------(2)
now, from the first equation find the value of $t$. We can write the first equation as
$
   \Rightarrow \sqrt 3 x - y - 4\sqrt 3 t = 0 \\
   \Rightarrow 4\sqrt 3 t = \sqrt 3 x - y \\
  \therefore t = \dfrac{{\sqrt 3 x - y}}{{4\sqrt 3 }} \\
 $
Now, putting the value of $t$ in second equation we get
$
   \Rightarrow \sqrt 3 tx + ty - 4\sqrt 3 = 0 \\
   \Rightarrow \sqrt 3 x\left( {\dfrac{{\sqrt 3 x - y}}{{4\sqrt 3 }}} \right) + \left( {\dfrac{{\sqrt 3 x - y}}{{4\sqrt 3 }}} \right)y - 4\sqrt 3 = 0 \\
   \Rightarrow \dfrac{{3{x^2} - \sqrt 3 xy + \sqrt 3 xy - {y^2} - 48}}{{4\sqrt 3 }} = 0 \\
   \Rightarrow 3{x^2} - {y^2} - 48 = 0 \\
 $
Now, converting the above equation of hyperbola in standard form we get,
$
  \dfrac{{3{x^2}}}{{48}} - \dfrac{{{y^2}}}{{48}} = 1 \\
  \dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{{48}} = 1 \\
 $
Comparing this equation with the standard equation $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$. We get
${a^2} = 16$ and ${b^2} = 48$
Now, we have to apply the above given relation between $a$, $b$ and $e$.
$ \Rightarrow {b^2} = {a^2}\left( {{e^2} - 1} \right)$
By putting the value of ${a^2}$ and ${b^2}$ in the above equation we get,
$
   \Rightarrow 48 = 16\left( {{e^2} - 1} \right) \\
   \Rightarrow \left( {{e^2} - 1} \right) = \dfrac{{48}}{{16}} \\
   \Rightarrow \left( {{e^2} - 1} \right) = 3 \\
   \Rightarrow {e^2} = 3 + 1 = 4 \\
   \Rightarrow e = \sqrt 4 \\
  \therefore e = \pm 2 \\
 $
Thus, the eccentricity of the given hyperbola is $2$.

Hence, option (B) is correct.

Note: The eccentricity for a hyperbola is always greater than one.
Similarly, we can find the eccentricity of ellipse by using the relation between $a$, $b$ and $e$ for ellipse that is ${b^2} = {a^2}\left( {1 - {e^2}} \right)$.
The eccentricity for an ellipse is always less than one.