JEE Advanced Model Paper 9 with solutions for online practice

Any point on the given hyperbola is (4t , 4/t)
Normal at this point is y - 4/t = t2 (x - 4t)
If the normal passes through P(h , k),then k - 4/t = t2 (h - 4t)
4t4−t3h + tk - 4 = 0
This equation has roots t1,t2,t3,t4 which are parameters of the four feet of normals on the hyperbola.Therefore,
Σt1 = h/4
Σt1t2 = 0
Σt1t2t3 = k/4 and
t1t2t3t4 = - 1
Therefore , t1−1+t2−1 + t3−1+t4−1 = k/4
y1+y2+y3+y4 = k
According to the question t12+t22+t32+t42 = h2/16 = k
Hence, the locus of (h , k) is x2 = 16 y