Given the ellipse with equation `x^(2) + 4y^(2) = 16`, find the length of major and minor axes, eccentricity, foci and vertices.

Explanation

First, we will write the equation in standard form by dividing by 64 and we get :
`(x^(2))/(16) + (y^(2))/(4) = 1`
i.e., `(x^(2))/((4^(2))) + (y^(2))/((2^(2)) = 1`
`rArr` a = 4 and b = 2
Since the denominator of `x^(2)` is larger, therefore the major axis is along x-axis and the minor axis along y-axis
(1) Length of major and minor axis are 8 and 4 respectively.
(2) Eccentricity i.e., `e = (c)/(a)`
Now `c = sqrt(a^(2) -b^(2)) " " ...(i)`
Dividing (i) by 'a', we get :
`rArr e = (sqrt(a^(2)-b^(2)))/(a)`
` rArr e = (sqrt(16-4))/(4)`
` rArr e = (2sqrt(3))/(4)`
`rArr e = (sqrt(3))/(2)`
(3) The co-ordinates of the foci are (c, 0) and (-c, 0) where `c = sqrt(a^(2) - b^(2)) = sqrt(16-4) = sqrt(12) = 2sqrt(3)`

`therefore (2sqrt(3), 0) and (-2sqrt(3), 0)` are the required co-ordinates
(4) The co-ordinates of he vertices are (a, 0) and `(-a, 0)` i.e., (4, 0) and `(-4, 0)` .