Flag Analytical Geometry> Locus of centroid...
question mark

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.

Jackie kumar kumar , 15 Years ago
Grade 11
anser 1 Answers
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

          

Provide a better Answer & Earn Cool Goodies

Enter text here...
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free

Enter text here...