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