Question

State whether the given statement is True or False.
The locus of the point of intersection of lines \(\sqrt{3} x−y−4 \sqrt{3} k=0\) and \(\sqrt{3} kx+ky−4\sqrt{3} =0\) for different value of \(k\) is a hyperbola whose eccentricity is \(2\).

Answer 1

We have lines
\(\sqrt{3}x−y−4\sqrt{3} k=0\ldots(1)\)
And \(\sqrt{3} kx+ky−4\sqrt{3} =0\ldots(2)\)
\(\Rightarrow k=\dfrac{\sqrt{3}x−y}{4\sqrt{3}}\) [from equation \((1)\)]
Put the value of k in equation \((2)\)
\(\Rightarrow (\sqrt{3} x+y) \times \dfrac{\sqrt{3}x−y}{4\sqrt{3}}=4\sqrt{3}\)
\(\Rightarrow 3x^2−y^2=48\)
\(\Rightarrow \dfrac{x^2}{16}−\dfrac{y^2}{48}=1\ldots(3)\)
Which is equation of a hyperbola

By comparing the equation \((3)\) with standardequation of hyperbola \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\)
we get
\(a^2=16\) and \(b^2=48\)
\(\Rightarrow a = 4 ,b=\sqrt{48}\)
\(\because e=\dfrac{\sqrt{a^2+b^2}}{a}\)
\(\Rightarrow e=\dfrac{\sqrt{16+48}}{4}~~~~~[\because a=4 ,b=\sqrt{48}]\)
\(\Rightarrow e=2\)

Hence, the given statement is true.