Question

An equation of an ellipse is given. y2 = 1 − 3x2 (a) Find the vertices, foci, and eccentricity of the ellipse. vertex (x, y) = (smaller y-value) vertex (x, y) = (larger y-value) focus (x, y) = (smaller y-value) focus (x, y) = (larger y-value) eccentricity (b) Determine the length of the major axis. Determine the length of the minor axis.

Answer 1

Explanation:

Given

`y^2=1-3x^2`

`3x^2+y^2=1`

`(x^2)/(((1)/(√(3)))^2)+(y^2)/(1)=1`

therefore it is a vertical ellipse

thus a=1

`b=(1)/(√(3))`

eccentricity of Ellipse

`e^2=1-(b^2)/(a^2)`

`e^2=1-(1)/((√(3))^2)`

`e^2=1-(1)/(3)`

`e^2=(2)/(3)`

`e=\sqrt{(2)/(3)}`

Focii are
`(0,ae)` and
`(0,-ae)`

`ae=1* \sqrt{(2)/(3)}`

thus focii are
`(0,\sqrt{(2)/(3)})` &
`(0,-\sqrt{(2)/(3)})`

(b) Length of major axis
`=2a=2* 1`

length of minor axis
`=2b=2* \sqrt{(2)/(3)}=2\cdot \sqrt{(2)/(3)}`