From any point P on the hyperbola x 2 /a 2 - y 2 /b 2 =1,three normals other than that at P are drawn.Find locus of the centroid of triangle formed by their feet.

							
let p be ( a secØ , b tanØ )
let a point on hyperbola be (h,k)
so  ,  h^2 /a^2    -  k^2 /b^2  = 1 ...........................(1)
      eq of normal at this point  ,
                       a^2 x / h  + b^2 y /k =   a^2 + b^2 .........................(2)
since this normal passes trough P,
                   a^3 secØ / h  + b^3 tanØ /k =  a^2 +b^2  ..........................(3)
let    a^3 secØ = A ,    b^3 tanØ  = B   ,  a^2 +b^2 = C,......then
        A/h + B/k = C.....................................(4)
from eq 4 we get,
      h = Ak/(ck-B),,,.........put ineq.  (1)
we get
      a^2 * C^2 * k^4  -  2 * C*B * k^3 +  (........)   k^2 + (....)  k +  const. = 0
   from this eq .
              k1 +k2 +k3 +k4 =  - ( -  2 * C*B)/   a^2 * C^2                      {   k1 +k2 +k3 +k4 = - b/a}
                                        =  2B/C
                                        = 2  b^3 tanØ/ ( a^2 + b^2)
                    since  ( h4,k4) is point P,    k4 = btanØ
                           k1 +k2 +k3  =   2  b^3 tanØ/ ( a^2 + b^2)    -  btanØ
                                               = -  btanØ * (a^2- b^2)/( a^2 + b^2) 
 
 for centroid ,            y c=  ( k1 +k2 +k3  )/3
                                      =    -  btanØ * (a^2- b^2) / 3( a^2 + b^2) 
in asimilar way we can get,
                            x c =  ( h1 +h2+ h3) /3
                                  = a secØ * (a^2- b^2) / 3( a^2 + b^2) 
if  (a^2- b^2) / 3( a^2 + b^2)  = G
    (xc, yc )  =   aG secØ  ,   -  b  GtanØ 
 which emplies that the locus of the centrid will be also a hyperbola, 
                           x^2 / (aG)^2    -    y^2 / (bG)^2   =1