If the length of the major axis of an ellipse | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the length of the major axis be $2a$ and the length of the minor axis be $2b$.Given that the length of the major axis is three times the lengt

Vedclass · Accepted Answer

$\frac{2\sqrt{2}}{3}$

Vedclass · Answer

(D) Let the length of the major axis be $2a$ and the length of the minor axis be $2b$.Given that the length of the major axis is three times the length of the minor axis:$2a = 3(2b) \implies a = 3b$.We know the formula for eccentricity $e$ of an ellipse is $e = \sqrt{1 - \frac{b^2}{a^2}}$.Substituting $a = 3b$ into the formula:$e = \sqrt{1 - \frac{b^2}{(3b)^2}}$$e = \sqrt{1 - \frac{b^2}{9b^2}}$$e = \sqrt{1 - \frac{1}{9}}$$e = \sqrt{\frac{8}{9}}$$e = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.Thus,the eccentricity is $\frac{2\sqrt{2}}{3}$.

If the length of the major axis of an ellipse is three times the length of its minor axis,then its eccentricity is

Similar Questions

The sum of focal distances of any point on the ellipse with major and minor axes as $2a$ and $2b$ respectively,is equal to

The equation of the ellipse having a vertex at $(6,1)$,a focus at $(4,1)$ and the eccentricity $e = \frac{3}{5}$ is

The locus of the point of intersection of perpendicular tangents to the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is:

The length of the latus-rectum of the ellipse,whose foci are $(2,5)$ and $(2,-3)$ and eccentricity is $\frac{4}{5}$,is

For an ellipse with eccentricity $e = \frac{1}{2}$,the centre is at the origin. If one of its directrices is $x = 4$,then the equation of the ellipse is

Vedclass Test Series

Exam Paper Generator

Online Exam Module