Get Answers to all your Questions

header-bg qa

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.

Answers (1)

True  

 Given equation of lines are \sqrt{3}x-y-4\sqrt{3}k=0

\sqrt{3}kx+ky-4\sqrt{3}=0

k=\frac{\sqrt{3}x-y}{4\sqrt{3}} andk=\frac{4\sqrt{3}}{\sqrt{3}x+y}

\frac{\sqrt{3}x-y}{4\sqrt{3}} =\frac{4\sqrt{3}}{\sqrt{3}x+y}

3x2-y2=48

\frac{x^{2}}{16}-\frac{y^{2}}{48}=1 which is the equation of hyperbola 

a2=16 and b2=48

e2=1+48/16=4

e=2

Posted by

infoexpert22

View full answer