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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' isa) (1-e)x2+(1+e)y2=a2b)(1+e)x2+(1-e)y2=a2c2c)(1+e)x2+(1-e)y2=a2d)(1-e)x2+(1+e)y2=a2e2(1-e)

147 Points
11 years ago

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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