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Let S and S' be the foci of the ellipse ...

Let S and S' be the foci of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` whose eccentricity is e. P is a variable point on the ellipse. Consider the locus the incentre of `DeltaPSS'`
. The locus of the incenter is a\an

A

a) ellipse

B

b) hyperbola

C

c) parabola

D

d) circle

Text Solution

Verified by Experts


Let the coordinates of P be `(a cos theta, b sin theta)`
Here, SP= Focal distance of point `P=a-ae cos theta`
`S''P=a+ae cos theta`
SS''=2ae
If (h,k) are the coordinates of the incenter of `DeltaPSS''`, then
`h=(2ae(a cos theta)+a(1-ecos theta)(-ae)+a(1+e cos theta)ae)/(2ae+a(1-e cos theta)+a(1+e cos theta)`
`as cos theta" "(1)`
`and k=(2ae(a sin theta)+a(1-ecos theta)xx0+a(1+e cos theta)xx0)/(2ae+a(1-e cos theta)+a(1+e cos theta))`
`=(eb sin theta)/((e+1))" "(2)`
Eliminating `theta` from (1) and (2), we get
`(x^(2))/(a^(2)e^(2))+(y^(2))/({be//(e+1)}^(2))=1`
Which clearly represents an ellipse. Let `e_(1)` be its eccentricity . Then
`(b^(2)e^(2))/((e+1)^(2))=a^(2)e^(2)(1-e_(1)^(2))`
or`e_(1)^(2)=1-(b^(2))/(a^(2)(e+1)^(2))`
`or e_(1)^(2)=1-(1-e^(2))/((e+1))=1-(1-e)/(1+e)`
`or e_(1)^(2)=(2e)/(e+1)`
`or e_(1)=sqrt((2e)/(e+1))`
Maximum area of reactangle is
`2(ae)(("be")/(e+1))=(2abe^(2))/(e+1)`
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