The locus of the point of intersection on the line 3 x-y-4 3 k=0 and 3 k x+k y-4 3 =0 for different real values of k is a hyperbola H . If e is the eccentricity of H , then 4 e^2=

Step-by-step solution

aligned & 3 x-y-4 3 k=0 & 3 k x+k y-4 3 =0 aligned Multiplying of k in Eq. (i) and adding in Eq. (ii), we get x=2 (1+k^2 ) k Multiplying of k in Eq. (i) and subtracting from Eq. (ii), we get aligned & y 3 =2 ( 1-k^2 k ) & x+ y 3 = 4 k & and x- y 3 =4 k & x^2- y^2 3 =16 aligned x^2 16 - y^2 48 =1 Now, e^2=1+ 48 16 =4 4 e^2=16