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

${a}^{2} - {b}^{2} = {c}^{2}$

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

${c}^{2} = 25 - 4$

${c}^{2} =21$

$c = \pm \sqrt{21}$

The foci:

$(\pm \sqrt{21} ,0)$

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

$\frac{ {x}^{2} }{ {b}^{2} } + \frac{ {y}^{2} }{ {a}^{2} } = 1$

$\frac{ {x}^{2} }{48} + \frac{ {y}^{2} }{64} = 1$