Question

Help with Ellipses please??

Answer 1

Write the equation of the ellipse 4x^2 + 9y^2 – 16x – 126y + 421 = 0 in standard form.

Answer 2

@jim_thompson5910

Answer 3

you need to complete the square for the x and y terms

Answer 4

I'll do the x terms and let you do the y terms

Answer 5

Thanks for helping :)

Answer 6

4x^2 + 9y^2 – 16x – 126y + 421 = 0

4x^2 – 16x + 9y^2 – 126y + 421 = 0

(4x^2 – 16x) + 9y^2 – 126y + 421 = 0

4(x^2 – 4x) + 9y^2 – 126y + 421 = 0

4(x^2 – 4x + 4 - 4) + 9y^2 – 126y + 421 = 0 ... See Note below

4( (x^2 – 4x + 4) - 4) + 9y^2 – 126y + 421 = 0

4( (x-2)^2 - 4) + 9y^2 – 126y + 421 = 0

4(x-2)^2 - 16 + 9y^2 – 126y + 421 = 0

4(x-2)^2 + 9y^2 – 126y + 421 - 16 = 0

4(x-2)^2 + 9y^2 – 126y + 405 = 0

I'll let you finish up and complete the square for the y terms

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Note: to complete the square for x^2 – 4x, you take half of the x coefficient -4 to get -2, then you square it to get 4.

You add and subtract this from x^2 – 4x to get x^2 – 4x + 4 - 4, which allows you to complete the square to go from x^2 – 4x to (x-2)^2 - 4

Answer 7

tell me what you get when you complete the square for the y terms

Answer 8

4(x-2)^2 + 9(y^2 – 14y) + 421 = 0

Sorry for taking too long....I got this

Answer 9

close

Answer 10

one sec

Answer 11

4( (x^2 – 4x + 4) - 4) you did this
so i'll do this too 9( (x^2 – 14x + 14) - 14)

Answer 12

is this right?

Answer 13

no, you take half of -14 to get -7

square it to get 49

Answer 14

so it *should* be this

9( (x^2 – 14x + 49) - 49)

Answer 15

oh, yeah, your right, that's why u did 4(x-2)^2, you got 2 cuz u took half of 4

Answer 16

exactly

Answer 17

so it's 9(y^2−14y+49)

Answer 18

so this is what you should get so far


4(x-2)^2 + 9y^2 – 126y + 405 = 0


4(x-2)^2 + (9y^2 – 126y) + 405 = 0


4(x-2)^2 + 9(y^2 – 14y) + 405 = 0


4(x-2)^2 + 9(y^2 – 14y + 49 - 49) + 405 = 0


4(x-2)^2 + 9( (y^2 – 14y + 49) - 49) + 405 = 0


4(x-2)^2 + 9( (y-7)^2 - 49) + 405 = 0


4(x-2)^2 + 9(y-7)^2 - 441 + 405 = 0

Answer 19

let me finish the rest ok

Answer 20

ok

Answer 21

4(x-2)^2 + 9(y-7)^2 - 441 + 405 = 0
4(x^2−4x+4)+9(y^2−14y+49)=−421 + 16 + 441

Answer 22

stick with 4(x-2)^2 + 9(y-7)^2 - 441 + 405 = 0

Answer 23

now add -441 and 405 to get ???

Answer 24

36

Answer 25

-36

Answer 26

so we then get

4(x-2)^2 + 9(y-7)^2 - 36 = 0

Answer 27

add 36 to both sides to get

4(x-2)^2 + 9(y-7)^2 = 36

Answer 28

your last step is to turn that 36 on the right side into a 1

how do you do this?

Answer 29

(x-2)^2/(9) + (y-7)^2/(4)= 1
right?????

Answer 30

perfect, you got it

Answer 31

and you're done since that's it in standard form

Answer 32

Yay!! Thanks for your help @jim_thompson5910

Answer 33

you're welcome