Single Choice

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$$\sqrt{3}$$
B2
Correct Answer
C$$\dfrac{2}{\sqrt{3}}$$
D$$\dfrac{4}{3}$$

Solution

Given, $$\sqrt{3}x - y - 4\sqrt{3}t = 0$$ and $$\sqrt{3}tx + ty - 4\sqrt{3} = 0$$
Eliminating $$t$$ from the given two equation,
from first equation we have,
$$t=\dfrac{\sqrt{3}x - y}{4\sqrt3}$$
And from second we have,
$$t=\dfrac{4\sqrt3}{\sqrt3 x+y}$$
we have $$ \dfrac{x^{2}}{16}-\dfrac{y^{2}}{48}=1 $$
whose eccentricity is $$ e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{1+\dfrac{48}{16}}=2 $$


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