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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
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best Sudarshan. Regards, Naga Ramesh IIT Kgp - 2005 batch
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