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There are exactly two points on the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`whose distances from its center are the same and are equal to `(sqrt(a^2+2b^2))/2dot`Then the eccentricity of the ellipse is`1/2`(b) `1/(sqrt(2))`(c) `1/3`(d) `1/(3sqrt(2))`A. `1//2`B. `1//sqrt(2)`C. `1//3`D. `1//3sqrt(2)` |
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Answer» (2) Since there are axactly two points on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` whosse distance from the center is the same, the poitns would either be the endpoints of the major axis or of the minor axis. But `sqrt((a^(2)+2b^(2)))//bgtb` So, the points are the vertices of the major axis hence, (1)`a=sqrt((a^(2)+2b^(2))/(2))` or `a^(2)=2b^(2)` `or e=sqrt(1-(b^(2))/(a^(2)))=(1)/(sqrt(2))` |
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