The length of the latus-rectum of the ellipse | vedclass

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

Question

(D) The distance between the foci is $2ae = \sqrt{(2-2)^2 + (5 - (-3))^2} = \sqrt{0^2 + 8^2} = 8$.Given the eccentricity $e = \frac{4}{5}$,we have $2a

Vedclass · Accepted Answer

$\frac{18}{5}$

Vedclass · Answer

(D) The distance between the foci is $2ae = \sqrt{(2-2)^2 + (5 - (-3))^2} = \sqrt{0^2 + 8^2} = 8$.Given the eccentricity $e = \frac{4}{5}$,we have $2ae = 8$,so $ae = 4$.Substituting $e = \frac{4}{5}$,we get $a 	imes \frac{4}{5} = 4$,which implies $a = 5$.For an ellipse,the relation between the semi-major axis $a$,semi-minor axis $b$,and the distance from the center to the focus $ae$ is $a^2 = b^2 + (ae)^2$ (where $a > b$).Here,$a = 5$ and $ae = 4$,so $5^2 = b^2 + 4^2$.$25 = b^2 + 16$,which gives $b^2 = 9$,so $b = 3$.The length of the latus-rectum is given by $\frac{2b^2}{a}$.Substituting the values,$	ext{L.R.} = \frac{2 	imes 9}{5} = \frac{18}{5}$.

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

Similar Questions

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

$A$ tangent is drawn to the ellipse $\frac{x^2}{32} + \frac{y^2}{8} = 1$ from the point $A(8, 0)$ to touch the ellipse at point $P$. If the normal at $P$ meets the major axis of the ellipse at point $B$,then the length $BC$ is equal to (where $C$ is the center of the ellipse) - ............ $units$

If the distance between the foci of an ellipse is $6$ and the distance between its directrices is $12$,then the length of its latus rectum is

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

The length of the latus rectum of the ellipse $\frac{x^2}{36} + \frac{y^2}{49} = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module