InterviewSolution
Saved Bookmarks
| 1. |
Let P be a poointon 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 |
|
Answer» <P> ae `R=("Product of all sides")/(4xx"Area of triangle")`ltbr Consider point `P(a cos theta, b SIN thea)` on the ELLIPSE. `:.` 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` |
|