Pratham Ashish
Last Activity: 15 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