Question

Does anyone know the minor axis vertices of 16x^2+4y^2+64 x+24y+36=0? The major axis vertices are (-2,-7),(-2,1), but what are the minor?

Answer 1

you would have to write it in the standard form of an ellipse in order to divine that knowledge I believe

Answer 2

or... input that equation into wolframalpha.com and read off the properties listed

Answer 3

I put it in Wolf and I only get the major axis vertices..

Answer 4

it gives a center; and a distance for minor and major

Answer 5

think of the center as the midpoint of the distance; in other words, divide it in half and use that value to determine the points for the vertexes

Answer 6

Oh well I had been putting in "minor axis vertices of" then my problem, not just my problem.. When I do that instead I get what I put in my attachment (along with the other things). I know 2 of these have to be the minor....

Answer 7

center (-2,-3) ; major 4 (4/2=2) x related; minor 2 (2/2=1) y related

major vertexes (-2, -3+2), (-2,-3-2)
minor verteces (-2+1,-3), (-2-1,-3)

if i see it right

Answer 8

Oops I never attached :P

Answer 9

notice have the same "x value" which makes the major axis (x related), have the same x values

Answer 10

notice the foci have the same ...

Answer 11

"minor vertices (-2+1,-3), (-2-1,-3)" how is there an equation in these..?

Answer 12

we need to measure out from the center (-2,-3) in order to get to our vertexes right?

Answer 13

Yeah,

Answer 14

im just writting out the math of it ....

Answer 15

-2+1 is the x component; -3 is the y component
-2-1 is the x component; -3 is the y component

Answer 16

wolfram says our minor is a length of 2; so we go out from the center +1 and -1

Answer 17

do you see why?

Answer 18

Kind of.

Answer 19

kind of is better than not at all :)

Answer 20

the math thing to do would be to "complete squares" for the x and y values and transform the equation into a standard: (x/a)^2 + (y/b)^2=1 format

Answer 21

16x^2 +4y^2 +64x +24y +36 = 0 ; reorganize

(16x^2 +64x +M) + (4y^2 +24y +N) +(36-M-N) = 0 ; reorganize

Answer 22

I kept trying and got (-4,-3),(0,-3)..?

Answer 23

(16x^2 +64x + ___) + (4y^2 +24y +___) = -36 + (___ + ___) ; factor

16(x^2 +4x +___) + 4(y^2 +6y +___) = -36 + (___+___); complete
squares

16(x^2 +4x +4) + 4(y^2 +6y +9) = -36 + (16(4)+4(9)); simplify

16(x+2)^2 + 4(y+3)^2 = -36 + 64+36; simplify some more

16(x+2)^2 + 4(y+3)^2 = 64 ; divide out 64


16(x+2)^2 4(y+3)^2 64
--------- + -------- = ----; simplify
64 64 64


(x+2)^2 (y+3)^2
------- + ------- = 1
4 16

the variable with the bigger denom is the major axis; in this case its y

the length of the major and minor axis are the sqrts of the denoms

length of major axis = sqrt(16) = 4

length of minor axis = sqrt(4) = 2

the center of the ellipse can be seen from negating the addons of the x and y

center (-2,-3)

if we start at the center and go along the major axis; want a length of 4 but start in the middle;

2 2
<........ (-2,-3) ........>

<....(-2,-5).....(center).....(-2,-1)....>

Answer 24

if we start at the center and go along the minor axis; we want a length of 2 but start in the middle;

1 1
<........ (-2,-3) ........>

<....(-3,-3).....(center).....(-1,-3)....>

Answer 25

Wow! That's a lot. But thank you :D

Answer 26

hopefully this dispels any confusion :)