The locus of the point of intersection of the lines √3kx+ky-4√3=0 and √3x-y-4√3k=0 is a co…

Question

The locus of the point of intersection of the lines $\sqrt3kx+ky-4\sqrt3=0$ and $\sqrt3x-y-4\sqrt3k=0$ is a conic, whose eccentricity is \_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

From the two equations, $k=\dfrac{4\sqrt3}{\sqrt3x+y}$ and $k=\dfrac{\sqrt3x-y}{4\sqrt3}$. Equating:
$$48=(\sqrt3x-y)(\sqrt3x+y)=3x^2-y^2.$$

So the locus is $\dfrac{x^2}{16}-\dfrac{y^2}{48}=1$, a hyperbola with $e=\sqrt{1+\dfrac{48}{16}}=\sqrt4=2$.

Solutions · 1