The eccentricity of the ellipse 4x^2 + 9y^2 = | vedclass

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

Question

(C) Given the equation of the ellipse: $4x^2 + 9y^2 = 36$. Dividing both sides by $36$,we get: $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Comparing this wit

Vedclass · Accepted Answer

$\frac{\sqrt{5}}{3}$

Vedclass · Answer

(C) Given the equation of the ellipse: $4x^2 + 9y^2 = 36$. Dividing both sides by $36$,we get: $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 4$. The formula for eccentricity $e$ is $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting the values: $e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$. Thus,the correct option is $C$.

The eccentricity of the ellipse $4x^2 + 9y^2 = 36$ is

Similar Questions

The eccentric angles of the extremities of the latus recta of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are given by

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

The eccentric angle of a point on the ellipse $x^2+3y^2=6$ lying at a distance of $2$ units from its centre is

The area enclosed by the ellipse $4x^2 + 9y^2 = 1$ is . . . . . . sq. units.

If the latus rectum of an ellipse subtends a right angle at the centre of that ellipse,then the eccentricity of that ellipse is

Vedclass Test Series

Exam Paper Generator

Online Exam Module