Question

Help me finish this problem!! I will Fan & medal

Answer 1

4x^2 - 8x + 25y^2 + 150y + 129 = 0
4(x^2 - 8x) + 25(y^2 + 150y) + 129 = 0
4(x^2 - 8x) + 25(y^2 + 150y) = -129
4(x^2 - 8x + 16) + 25(y^2 + 150y + 5625) = -129
4(x^2 - 8x + 16) + 25(y^2 + 150y + 5625) = -129 + 4(16) + 25(5625)
4(x^2 - 8x + 16) + 25(y^2 + 150y + 5625) = 140560
4(x - 4)^2 + 25(y + 75)^2 = 140560
4(x - 4)^2/140560 + 25(y + 75)^2/140560 = 140560/140560
4(x - 4)^2/140560 + 25(y + 75)^2/140560 = 1

This is as far as i got.. i dont know how to do the rest ??

Answer 2

Write the standard form of the ellipse 4x^2 + 25y^2 - 8x + 150y + 129 = 0
This was the question^

Answer 3

its 7

Answer 4

Um no

Answer 5

um yes

Answer 6

How do you get that? Do you know what standard form is?

Answer 7

4x^2 - 8x + 25y^2 + 150y + 129 = 0
4(x^2 - 8x) + 25(y^2 + 150y) + 129 = 0 <--- when you factor 25 from the y terms... don't forget to divide 150 by 25

Answer 8

Im sorry i forgot to say this is for ellipse

Answer 9

What do you mean?

Answer 10

I mean this step is wrong:
\[ 25y^2 +150y = 25(y^2+150y) \]

Answer 11

if you "distribute" the 25 in
25(y^2 +150y) you get 25y^2 + 3750y
which is not the original

Answer 12

Im confused haha :/ im just following what my notes say it doesnt show you have to multiply 25 by 150

Answer 13

I was completing the square

Answer 14

1/2 * 150 = 75 = 75^2
(y^2 + 150y + 5625)

Answer 15

yes. And the first step is factor... but you are missing an idea.
look at the x's
4x^2 - 8x

you can think of that as 4 * x^2 - 4* 2x
and factor out a 4: 4(x^2 - 2x)
(and to show this is correct, we can "undo" the factoring by distributing: multiply 4 times each term inside the parens and you get 4x^2 -8x which is what we started with...

the point of all that is 4x^2 - 8x becomes 4(x^2 -2x)

compare that to what you got (on your second line)

Answer 16

**fixed the typo's with a minus sign

Answer 17

ohhh so i went wrong on my second step?

Answer 18

yes, can you re-do the second line ?

Answer 19

it should be :
4x^2 - 8x + 25y^2 + 150y + 129 = 0

4(x^2 - 2x) + 25(y^2 + 6y) + 129 = 0

Answer 20

yes. now the next step

Answer 21

it quals swwag



(worth t)

Answer 22

Wow i did a worksheet and skipped this step on all of them :// lol
completing the square?

Answer 23

4(x^2 - 2x + 1) + 25(y^2 + 6y + 9) = -129

Answer 24

close.

Answer 25

but there is this idea: we can add 0 to the equation and not change it.
when you changed
4(x^2 - 2x) to 4(x^2 - 2x+1) that makes it a different equation
but if we add 0 (written in the peculiar way 1 - 1 ) it's ok:
4(x^2 - 2x +1 - 1)

Answer 26

the same for the y's
you want
\[ 25(y^2 + 6y + 9 - 9) \]

Answer 27

for the x stuff:
4(x^2 - 2x +1 - 1)

the idea is that x^2-2x+1 is the same as (x-1)^2
so we have
\[ 4( (x-1)^2 -1 ) \]

Answer 28

Im confused why you would do that..? Becuase you have to complete the square and in my notes it doesnt show that it just shows completing the square as i did

Answer 29

it depends on the steps they are teaching you. The ideas are the same, but the details might be different.

Here is the simplest problem:

x^2 -2x = 0

you look at the middle number (-2), divide it in half (to get -1). square it to get -1*-1=+1
now add +1 to both sides of the equation
x^2 -2x +1 = 0+1 (when we add 1 to the left side, to keep things balanced we add 1 to the right side)

finally re-write as
(x-1)^2 = 1

Answer 30

another way (almost the same, really ) is
x^2 -2x +1 - 1= 0
we add +1 (to complete the square) and -1 (to compensate for adding the +1)
now rewrite as
(x^2 -2x +1) - 1= 0
(x-1)^2 -1 = 0
or (if we add +1 to both sides:
(x-1)^2 = 1
just like before

Answer 31

Oh okay i see maybe they just dont show that step but i got my answer using 4(x^2 - 2x + 1) + 25(y^2 + 6y + 9) = -129
this

Answer 32

yes. And that is half-way correct. For example, with the x's you added +1 to complete the square. But we have to keep things balanced. The easiest way (for me) is add -1 at the same time .... so you should write the x part as
4(x^2 - 2x + 1 - 1)

Answer 33

But then wouldnt it just go back to 4(x^2 - 2x) ? or is it the same thing

Answer 34

and then you couldnt factor it to (x - 1)^2

Answer 35

it could (proving we did not change the value), but we could put parens in like this
\[4\left((x^2 - 2x + 1 )-1 \right) \]

Answer 36

now re-write x^2-2x+1 as (x-1)^2
\[ 4 \left( (x-1)^2 -1\right) \]

Answer 37

finally, distribute the 4:
\[ 4 \left( (x-1)^2 -1\right) \\ = 4(x-1)^2 -4 \]

Answer 38

and we are making progress. Can you do the y's ?

Answer 39

Ughh hahaha this is confusing me

Answer 40

then go step by step

Answer 41

for the y's
we started with
25y^2 + 150y

factor out 25 (because we want a y^2 by itself when we complete the square)
we get
25(y^2 + 6y)

so far so good, ?

Answer 42

Yeha

Answer 43

next step: to complete the square for y^2+6y we must add +9
to compensate (this part confuses you, but go with the flow...) also add -9:
25(y^2 + 6y+9 - 9)

Answer 44

Okay haha

Answer 45

if you don't understand about the -9, we will get back to it... after we finish this dumb question .

25(y^2 + 6y+9 - 9)

next step: we know that y^2 + 6y+9 can be rewritten as (y+3)^2 so let's do that:

\[ 25\left( (y+3)^2 -9 \right) \]

Answer 46

I'll let you do the next step. Can you distribute the 25 in
25( (y+3)^2 -9 )
?

Answer 47

25 * -9?

Answer 48

25 times each thing inside the parens and then drop the parens

Answer 49

think of the parens as a "package" and you have 25 of each thing inside the package

Answer 50

like 25 happy meals... that means you have 25 of each thing in the happy meal

Answer 51

like this
\[ 25( (y+3)^2 -9 ) \\
25(y+3)^2 -9\cdot 25 \\ 25(y+3)^2-225\]

Answer 52

okay so far so good

Answer 53

ok. Let's put the pieces together. we now have
\[ 4(x-1)^2 -4 + 25(y+3)^2-225 = -129 \]

I would combine all the "pure numbers"
can you do that ?

Answer 54

yes so you would have 225 -129 = 96

Answer 55

don't forget the -4 on the left side

Answer 56

so 100

Answer 57

OH!!! so it works both ways :)

Answer 58

Your way may be more simple

Answer 59

now divide both sides by 100 (standard equation of an ellipse has 1 on the right side)

Answer 60

you get 25 on the left and 4 on the right same answer i got :)

Answer 61

you should get
\[ 4(x-1)^2 + 25(y+3)^2 = 100 \\
\frac{(x-1)^2}{25} + \frac{(y+3)^2}{4} = 1 \\
\frac{(x-1)^2}{5^2} + \frac{(y+3)^2}{2^2} = 1
\]

an ellipse with its center at (1,-3) and semi-major axis 5 and semi-minor axis 2

Answer 62

4(x^2 - 2x + 1) + 25(y^2 + 6y + 9) = -129
after this step i did
4(x^2 - 2x + 1) + 25(y^2 + 6y + 9) = -129 + 4(1) + 25(9)

Answer 63

Yeah thats what i go only a^2 and b^2 on my multiple choice were 25 under left denominator and 4 under right

Answer 64

But anyways I understand now!! :)

Answer 65

****
4(x^2 - 2x + 1) + 25(y^2 + 6y + 9) = -129
after this step i did
4(x^2 - 2x + 1) + 25(y^2 + 6y + 9) = -129 + 4(1) + 25(9)
****

ok, but that really should be *one* step... After each step, the equation should still be true.
when you put in the +1 (for the x) you can compensate by adding 4*1 on the right side
when you put in the +9 for the y, you can compensate by add 9*25 on the right side.

So if you wrote it as one step, it would be more correct.

But so long as you can do it (anyway that works!) is ok