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Vaibhav Mathur Grade: 12
        

If P is any point lying on the ellipse x2/a2 +y2/b2=1,a>b with foci S and S',then the locus of the incentre of the triangle PSS' is


a) (1-e)x2+(1+e)y2=a2


b)(1+e)x2+(1-e)y2=a2c2


c)(1+e)x2+(1-e)y2=a2


d)(1-e)x2+(1+e)y2=a2e2(1-e)

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear Vaibhav


x2/a2 +y2/b2=1


S=(ae,o)


S' =(-ae,o)


let a point P =(a cosΘ,bsinΘ)


we now focul distance SP=a-ex=a-eacosΘ


and  S'P = a+ex = a+excosΘ


and SS'=2ae


so incenter of triangle x=(SS' acosΘ + SP (-ae) +S'P(ae))/( SP +S'P+SS')


                                x=aecosΘ


simply for y co ordinate  y= e(bsinΘ)/(1+e)


so cosΘ =x/ae


and sinΘ =y(1+e)/eb


square and add locus will come (1-e)x2+(1+e)y2=a2e2(1-e)


 


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Badiuddin

7 years ago
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