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Let P be a pooint on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1(agtb)` in the first or second quadrants whowse foci are `S_(1)and S_(2)`. Then the least possible value of circumradius of `DeltaPS_(1)S_(2)` will be
A. ae
B. be
C. `(ae)/(b)`
D. `(ae^(2))/(b)`

1 Answer

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Best answer
We known that circumradius of triangle
`R=("Product of all sides")/(4xx"Area of triangle")` ltbr Consider point `P(a cos theta, b sin thea)` on the ellipse.
image
`:.` Circumrdius , `R=(a(1-ecostheta)xxa(1+ecostheta)xx2ae)/(4xx(1)/(2)b sin thetaxx2ae)`
`=(a^(2)(1-e^(2)cos^(2)theta))/(2b sin theta)`
`=(a^(2)(1-e^(2)(1-sin^(2)theta)))/(2b sin theta)`
`=(a^(2))/(2)(e^(2)sintheta+(b^(2))/(a^(2))"cosec"theta)`
`ge(a^(2))/(2b)xx(2be)/(a)ae`
`:.R_("min")=(a^(2))/(2b)xx(2be)/(a)=ae`

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