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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. 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.
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
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
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