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# given the base and sum of the other two sides of triangle,prove that locus of its incentre is an ellipse. Grade:Upto college level

## 1 Answers

11 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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