The distance between the directrices of the e | vedclass

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

Question

(C) Given the equation of the ellipse: $\frac{x^2}{36}+\frac{y^2}{20}=1$.Comparing this with the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(C) Given the equation of the ellipse: $\frac{x^2}{36}+\frac{y^2}{20}=1$.Comparing this with the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we have $a^2=36$ and $b^2=20$,which gives $a=6$.The eccentricity $e$ is calculated as $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{20}{36}} = \sqrt{\frac{16}{36}} = \frac{4}{6} = \frac{2}{3}$.The equations of the directrices for an ellipse with $a > b$ are $x = \pm \frac{a}{e}$.Substituting the values,we get $x = \pm \frac{6}{2/3} = \pm 9$.The distance between the directrices is the difference between these two values: $|9 - (-9)| = 18$.

The distance between the directrices of the ellipse $\frac{x^2}{36}+\frac{y^2}{20}=1$ is

Similar Questions

If $\alpha$ and $\beta$ are the eccentric angles of the extremities of a focal chord of an ellipse,then the eccentricity of the ellipse is

If $4x + 2y + n = 0$ is a normal to the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$,then $n = $

If the distance between the directrices of an ellipse is three times the distance between its foci,then the eccentricity of the ellipse is:

The eccentricity of the ellipse $25x^2 + 16y^2 = 400$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module