IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

sarthakvij

prove that locus of point of intersection of the lines

3x-y-4

3k=0 and

3kx+ky-4

3=0 , for diffrent values of k is a hyperbola whose eccentricity is 2.

-square root

puneet

hiiiii

Let us try to find the point of intersection of the two lines ..

SO let us solve

3x-y-4

3k = 0

and,

3kx+ky-4

3 = 0 simlutaneously

Solving we get x = 2(k² + 1)/k

and y = 2

3(k² - 1)/k

and, now if we find (x)² - (y/

3)²

we get (x)² - (y/

3)² = 16

so, x²/16 - y²/48 = 1

This is equation of hyperbola and the eccentrity is clearly

(48/16 + 1) = 2

I hope u have got the question ...

cheers

sarthakvij

thanks sir