Ramesh V
Last Activity: 15 Years ago
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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Naga Ramesh
IIT Kgp - 2005 batch