 # given the base and sum of the other two sides of triangle,prove that locus of its incentre is an ellipse. 12 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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