Question

Hhhhhelp plzzz look at the photo answer all three if possible

Answer 1

QUESTION 1

Given foci to be


`(\pm1,0)`


`c = \pm1`

and

co-vertices


`(0, \pm2)`


`b = \pm \: 2`


`{b}^(2) = 4`

This implies the minor axis of the ellipse is on the y-axis and the major axis is on the x-axis.

We use the equation


`{a}^(2) - {b}^(2) = {c}^(2)`

to determine the vertices.


`{a}^(2) - {( \pm2)}^(2) = { ( \pm1)}^(2)`


`{a}^(2) - 4 = 1`


`{a}^(2) = 5`

The equation of the ellipse is given by;


`\frac{ {x}^(2) }{ {a}^(2) } + \frac{ {y}^(2) }{ {b}^(2) } = 1`


`\frac{ {x}^(2) }{ 5} + \frac{ {y}^(2) }{4} = 1`

QUESTION 2

The given ellipse has

Foci


`(0,\pm2)`


`c = \pm2`

and co-vertices,


`(\pm1,0)`


`b = \pm1`


`{b}^(2) = 1`

We use the equation,


`{a}^(2) - {b}^(2) = {c}^(2)`

to determine the vertices.


`{a}^(2) - { (\pm1)}^(2) = {( \pm2)}^(2)`


`{a}^(2) - 1= 4`


`{a}^(2) =5`

The major axis of the ellipse is on the y-axis this time.

The equation is given by;


`\frac{ {x}^(2) }{ {b}^(2) } + \frac{ {y}^(2) }{ {a}^(2) } = 1`


`\frac{ {x}^(2) }{ 4} + \frac{ {y}^(2) }{ 5} = 1`

QUESTION 3

The given ellipse has

Foci


`(0,\pm4)`

and co-vertices,


`(\pm4,0)`


`b = \pm4`


`{b}^(2) = 16`


`{a}^(2) - {b}^(2) = {c}^(2)`


`{a}^(2) - 16= 16`


`{a}^(2) = 32`

The major axis is on the y-axis.

The equation is,


`\frac{ {x}^(2) }{ {b}^(2) } + \frac{ {y}^(2) }{ {a}^(2) } = 1`


`\frac{ {x}^(2) }{16} + \frac{ {y}^(2) }{32 } = 1`