given the base and sum of the other two sides of triangle,prove that locus of its incentre is an ellipse.

							
Since the base lenght is given and sum of 2 sides is aslo given, lets take the given triangle as follows from an ellipse
where the base is between 2 focii and the other side is P(x,y) a point on ellipse
the 3 points are A(-ae,0)  B(ae,0)   P(x,y)
Lets take given values as                                   base=2ae     [ e is eccentricity of ellipse ]
and                           sum of other 2 sides (PA+PB) =2a 

From properties of ellipse, PB=ePZ  = e(a/e -x) = a-ex
                                                  PA=ePZ  = a+ex  
                                                  AB=2ae
Let I(h,k) be the locus of incentre of triangle ABP.
h = (2aex +(a+ex)ae -ae(a-ex) )/ (2a +2ae)
 h = ex
k = 2aey / 2a(1+e)
k = ey / (1+e)
i.e., x=h/e and y=k(1+e)/e
as we have x2/a2 +y2/b2 =1
                        i.e.,                       h2/a2e2 +k2(1+e)2/a2e2(1-e2) =1
now in the above equation, to prove this is form ELLIPSE   [ NOTE: here I(h,k) is the locus of incentre of triangle ABP.]
put a' =ae
let  (1-e2)/(1+e)2 = 1-e' 2
i.e.,   e' 2= 1 - (1-e2)/(1+e)2 = 2e/(1+e)
now  e' 2= 2e/(1+e) = (e+e)/(1+e) < 1       (as eccentricity of ellipse e<1)
hence e' <1
hence from above we can write in form of ELLIPSE as
h2/(a')2 + k2/(a')2(1-(e')2) =1  which is ellipse
Hence, the problem solved
 
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IIT Kgp - 2005 batch