If eccentricity of an ellipse be 5/8 and distance between its foci be 10 , then its latus rectum is

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

easy

If the eccentricity of an ellipse be $5/8$ and the distance between its foci be $10$, then its latus rectum is

A

$39/4$

B

$12$

C

$15$

D

$37/2$

Solution

(a) $a = \frac{{10}}{{\frac{{2.5}}{8}}} = 8$, $b = 8\sqrt {1 – \frac{{25}}{{64}}} = 8\frac{{\sqrt {39} }}{8}$ = $\sqrt {39} $

Latus rectum$ = \frac{{2{b^2}}}{a} = \frac{{2 \times 39}}{8} = \frac{{39}}{4}$.

Standard 11

Mathematics

Share

Similar Questions

The eccentricity of ellipse $(x-3)^2 + (y -4)^2 = \frac{y^2}{9} +16 ,$ is –

normal

The equation of the tangent to the ellipse ${x^2} + 16{y^2} = 16$ making an angle of ${60^o}$ with $x$ – axis is

medium

The locus of the mid point of the line segment joining the point $(4,3)$ and the points on the ellipse $x^{2}+2 y^{2}=4$ is an ellipse with eccentricity

hard

(JEE MAIN-2022)

If a number of ellipse be described having the same major axis $2a$ but a variable minor axis then the tangents at the ends of their latera recta pass through fixed points which can be

normal

An ellipse, with foci at $(0, 2)$ and $(0, -2)$ and minor axis of length $4$, passes through which of the following points?

hard

(JEE MAIN-2019)