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

Answers : (1)

Ramesh V
70 Points
										

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


7 years ago
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