If the latus rectum of an ellipse is equal to | vedclass

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

Question

(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. Length of the latus rectum $= \frac{2b^2}{a}$. Length of

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. Length of the latus rectum $= \frac{2b^2}{a}$. Length of the minor axis $= 2b$. According to the given condition,the latus rectum is equal to half of the minor axis: $\frac{2b^2}{a} = \frac{1}{2} (2b) = b$. Dividing both sides by $b$ (since $b 
eq 0$),we get $\frac{2b}{a} = 1$,which implies $\frac{b}{a} = \frac{1}{2}$. Squaring both sides,we get $\frac{b^2}{a^2} = \frac{1}{4}$. The eccentricity $e$ of an ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting the value of $\frac{b^2}{a^2}$: $e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

If the latus rectum of an ellipse is equal to half of its minor axis,then its eccentricity is

Similar Questions

The equation of an ellipse in its standard form,given the distance between its foci is $2$ units and the length of its latus rectum is $\frac{15}{2}$ units,is

The equation of the ellipse whose one focus is at $(4, 0)$ and whose eccentricity is $4/5$,is

The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are:

If the length of the minor axis of an ellipse is equal to one-fourth of the distance between the foci,then the eccentricity of the ellipse is:

The length of the latus rectum of the ellipse $9x^2 + 4y^2 = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module