Question

Right in equation of the ellipse The rest of the question is in the photo can you plz answer both

Answer 1

QUESTION 1

The given ellipse has equation:


`4 {x}^(2) + 25 {y}^(2) = 100`

Divide through by 100 to get;


`\frac{ {x}^(2) }{25} + \frac{ {y}^(2) }{4} = 1`

The ellipse has it major axis on the x-axis.


`{a}^(2) = 25`


`a = \pm5`

The vertices is

(-5,0) and (5,0).

Also


`{b}^(2) = 4`


`b = \pm2`

The co-vertices are,

(0,-2), (0,2).

We use


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


`{c}^(2) = {( \pm5)}^(2) - {( \pm2)}^(2)`


`{c}^(2) = 25 - 4`


`{c}^(2) =21`


`c = \pm √(21)`

The foci:


`(\pm √(21) ,0)`

We plot all these points and graph our ellipse.

See attachment

QUESTION 2.

If the ellipse has a vertex at

(0,-8) and a focus at (0,4) then


`{a}^(2) = 64`

and


`{c}^(2) = 16`

we can use


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

to determine the value if b.


`{( - 8)}^(2) - {b}^(2) = {4}^(2)`


`64- {b}^(2) = 16`


`{b}^(2) = 64 - 16`


`{b}^(2) = 48`

The equation of the ellipse is


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


`\frac{ {x}^(2) }{48} + \frac{ {y}^(2) }{64} = 1`