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Prove that any point on the ellipse whose foci are`(-1,0)`and `(7,0)`andeccentricity is `1/2`is `(3+8costheta,4sqrt(3)sintheta),theta in Rdot` |
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Answer» The foci are (-1,0 ) and (7,0) . The distance between the foci is 2ae=8 or ae=4 Since e=1/2, we have a=8 Now, `b^(2)=a^(2)(1-e^(2))` or `b^(2)=48` or `b=4sqrt(3)`. The center of the ellipse is the midopoint of the joining two foci. Therefore, the coordinates of the center are (3,0) Hence its equation is `((x-3)^(2))/(8^(2))+((y-0)^(2))/((4sqrt(3)))=1 " "(1)` Thus, hte parameter coordinates of a point on (1) are `(3+8 cos theta, 4 sqrt(3)sin theta)` |
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