prove that the locus of the point of intersection of the lines x√3-y- 4k√3=0 and √3kx+ky-4√3=0 for different values of - Maths - Conic Sections

Question

prove that the locus of the point of intersection of the lines
x√3-y- 4k√3=0 and √3kx+ky-4√3=0 for
different values of k is a hyperbola whose e is 2

Neha Sethi · Accepted Answer

Dear student
The equations 3x-y-43k=0 and 3kx+ky-43=0 can be rewritten as3x-y=43k and 3kx+ky=43 respectively
Multipying the equations:3kx2-ky2=48k⇒3kx248k-ky248k=1⇒x216-y248=1This is the standard equation of a hyperbola , where a2=16 and b2=48Eccentricity, e=a2+b2a2⇒e=16+4816⇒e=84⇒e=2
Regards

prove that the locus of the point of intersection of the lines x√3-y- 4k√3=0 and √3kx+ky-4√3=0 for different values of k is a hyperbola whose e is 2.