Question

In an ellipse, the distance between its foci is 6 and minor axis is 8. then its eccentricity is

Answer 1

hola there

answer is in picture..
Hope you like it

Answer 2

Answer:

$e=\frac{3}{5}$

Step-by-step explanation:

Given :  The distance between its foci is 6 and minor axis is 8

To Find : eccentricity

Solution:

Equation : $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 (a>b)$

We are given 2b = 8  --1

And 2ae = 6   --2

With 1  and 2

$\frac{b}{ae}=\frac{4}{3}$

Squaring both sides

$\frac{b^2}{a^2}=\frac{16}{9}e^2$

we know that $b^2=a^2(1-e^2)$

$(1-e^2)=\frac{16}{9}e^2$

$e=\frac{3}{5}$

Hence  its eccentricity is $e=\frac{3}{5}$