# From any point P on the hyperbola x2/a2 - y2/b2 =1,three normals other than that at P are drawn.Find locus of the centroid of triangle formed by their feet.

Pratham Ashish
17 Points
14 years ago

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