Answer
$\dfrac{x^2}{4851}+\dfrac{y^2}{4900}=1$
Work Step by Step
The eccentricity of the ellipse $\dfrac{x^2}{m^2}+\dfrac{y^2}{n^2}=1$ when $m \gt n$ is given by:
$e=\dfrac{\sqrt {m^2-n^2}}{m}$
The foci of the ellipse are: $(\pm me,0)$ and the directrices are given as: $x=\pm \dfrac{m}{e}$
Given: $e=0.1$ ; vertices: $(0,\pm 70)$
Now, $c=(70)(0.1)=7$ and $n^2=m^2-c^2=4900-49=4851$
Here, the equation of the ellipse is:
$\dfrac{x^2}{4851}+\dfrac{y^2}{4900}=1$
