vertices of an ellipse are $(0,\pm 10)$ and its eccentricity $e=4/5$ then its equation is ;

Let the equation of the required ellipse be

$\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\longrightarrow \left( 1 \right)$

since the vertices of the ellipse are on $y$-axis, so the coordinate of the vertices are $\left( 0,\pm b \right)$

$\therefore b=10\\ Now,\quad { a }^{ 2 }=b^{ 2 }\left( 1-{ e }^{ 2 } \right) \\ \Rightarrow { a }^{ 2 }=100\left( 1-\dfrac { 16 }{ 25 } \right) \\ \Rightarrow { a }^{ 2 }=36\\$

substituting the value of ${ a }^{ 2 }$ and ${ b }^{ 2 }$ in equation $(1)$

we get, $\dfrac { { x }^{ 2 } }{ 36 } +\dfrac { { y }^{ 2 } }{ 100 } =1\\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }=3600\\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }-3600=0$