Question

Help please

Answer 1

compare this w/ your equation to determine the center (h,k)

Answer 2

(-3,2)

Answer 3

(x-h)^2 = (x-2)^2

h = ?

Answer 4

vertices are (2,0)(2,-6)

Answer 5

-3

Answer 6

+3

Answer 7

notice how

(x-2) = (x-h)

just comparing them side by side what do you think h might be equal to?

Answer 8

hint: subtract x from both sides

Answer 9

-2?

Answer 10

almost

-2 = -h

h = ?

Answer 11

2

Answer 12

good, let's repeat this process to find k

(y-k)^2 = (y+1)^2

k = ?

Answer 13

-1?

Answer 14

good

so (h,k) = (2,-1) = your center

Answer 15

was this part 3?

Answer 16

we are still on part I (the center.)

Answer 17

oh okay

Answer 18

now, find the value of a by comparing the standard equation to the equation in the problem

Answer 19

compare these side by side, what's the value of a?

Answer 20

36?

Answer 21

close but not quite

a^2 = 36, a = ?

Answer 22

6

Answer 23

good

so the coordinates of the vertices are: (h+a,k), (h-a,k)

plug in the h, a, and k values to get the coordinates

Answer 24

(2,0)(2,-6)

Answer 25

h = 2
k = -1
a = 6

re-try your calculations.

Answer 26

(8,-1)(-4,-1)

Answer 27

good

for part III) we will need to calculate the value of b

compare the equation in the problem w/ the standard form equation to find the value of b

Answer 28

(y+1)^2/25 = (y-k)^2/b^2 find b

Answer 29

if you just compare the denominators

25 = b^2

b = ?

Answer 30

5

Answer 31

good

now, we need to find c

c^2 = a^2 - b^2 for ellipses

c = ?

Answer 32

** should also mention, leave it in radical form

Answer 33

(2,2)(2,-8)

Answer 34

would this be the foci?

Answer 35

no

Answer 36

calculate c where c^2 = a^2 - b^2

as a reminder, we determined a to be 6 and b to be 5

Answer 37

36-25=11^2=22

Answer 38

not sure where you're getting 22 from but yes, c^2 = 11 so c = sqrt(11)

then the foci are just (h+c,k) and (h-c,k) so (2+sqrt(11), -1) and (2-sqrt(11),-1)

Answer 39

anyway, for IV

we determined the vertices to be (8,-1)(-4,-1)

major axis is along x, it's 2 * 6 = 12 units long
minor axis is along y, it's 2 * 5 = 10 units long

this should be sufficient to graph your ellipse

Answer 40

so it would be (12,0) and (0,10)

Answer 41

no

the ellipse is not centered at 0

12 and 10 are the lengths of the axes not the coordinates of the vertices.

Answer 42

create an ellipse with vertices (8,-1)(-4,-1), major axis length 12, minor axis length 10.

Answer 43

would it just be the equation given

Answer 44

it is asking for a graph not an equation

Answer 45

using the information provided, graph the ellipse, labelling the vertices.

Answer 46

check the minor axis again

if the center is (2,-1) then the ellipse crosses (2,-1+5) and (2,-1-5) so (2,4) and (2,-6)

Answer 47

good, that's about right