Question

What are the foci of the ellipse given by the equation 4x^2 + 9y^2 – 32x + 18y + 37 = 0?

Answer 1

do you know what the standard form for the equation of an ellipse is?

Answer 2

yes

Answer 3

can you write it please?

Answer 4

x^2/a^2 + y^2/b^2 =1
thats for an horizontal ellipse

X^2/b^2 + y^2/a^2 = 1
thats for a vertical ellpse

Answer 5

almost - the more general form is (for a horizontal ellipse):\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]where (h, k) are the coordinates of the center of the ellipse

Answer 6

so, what you need to do is to first convert your equation into this form

Answer 7

do you know how to do that?

Answer 8

yes but I think I got the wrong equation

Answer 9

if you write out your steps then I can help you spot where you may have gone wrong

Answer 10

i completed the squares and all that and I end up with the equation (x-4)^2/93 + 3(y+1)^2/124

Answer 11

ok - the equation you ended up with is wrong - if you write out each step you took then I can help you spot where you may have made a mistake

Answer 12

it is usually simpler to concentrate on just the x components, and then just the y components

Answer 13

so lets start with:\[4x^2-32x\]how can you complete the square here?

Answer 14

4(x^2 - 8 + 16)

Answer 15

\[4x^2-32x\ne4(x^2-8+16)\]

Answer 16

the first step would be to take out any common factors

Answer 17

so we start with:\[4x^2-32x=4(x^2-8x)\]

Answer 18

I meant - take out any constant factors (i.e. numeric values only)

Answer 19

now try and complete the square on:\[x^2-8x\]

Answer 20

to do this we halve the coefficient of x to get:\[(x-4)^2=x^2-8x+16\]therefore we know that:\[x^2-8x=(x-4)^2-16\]does that make sense?

Answer 21

yes

Answer 22

good, so we can now go back to the original components in x to get:\[4x^2-32x=4(x^2-8x)=4((x-4)^2-16)=4(x-4)^2-64\]agreed?

Answer 23

ok

Answer 24

great! now lets tackle the terms in y:\[9y^2+18y\]can you try to complete the square here using the same techniques as above please?

Answer 25

9(y+1) = 9

Answer 26

thats not correct - look carefully at how we worked with the x terms and try again please.

Answer 27

what is the common constant factor in:\[9y^2+18y\]

Answer 28

9

Answer 29

correct, so we can first simplify this to:\[9y^2+18y=9(y^2+2y)\]

Answer 30

next we try and complete the square for:\[y^2+2y\]

Answer 31

have a go at doing this...

Answer 32

dont u have Y^2 + 2y + (2/2)^2

Answer 33

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

Answer 34

I don't understand where you are getting the last line from?

Answer 35

9(y+1) = -9

Answer 36

where did this come from?

Answer 37

we are trying to complete the square for:\[y^2+2y\]

Answer 38

(Y + 1)^2 = 1

Answer 39

ok - what would you get if you expanded:\[(y+1)^2\]

Answer 40

y^2 + 2y + 1

Answer 41

correct, so we can therefore say that:\[y^2+2y=(y^2+2y+1)-1=(y+1)^2-1\]

Answer 42

ok

Answer 43

so, lets summarise what we have found so far...

Answer 44

\[4x^2-32x=4(x-4)^2-64\]\[9y^2+18y=9(y^2+2y)=9((y+1)^2-1)=9(y+1)^2-9\]agreed?

Answer 45

i agreed

Answer 46

now lets substitute these back into your original equation:\[4x^2+9y^2-32x+18y+37=0\]to get:\[4(x-4)^2-64+9(y+1)^2-9+37=0\]which simplifies to:\[4(x-4)^2+9(y+1)^2=36\]make sense so far?

Answer 47

yes

Answer 48

good - so now we divide both sides by 36 to get:\[\frac{(x-4)^2}{9}+\frac{(y+1)^2}{4}=1\]and, writing this in the standard form, we get:\[\frac{(x-4)^2}{3^2}+\frac{(y+1)^2}{2^2}=1\]

Answer 49

ok

Answer 50

you should now be able to find the coordinates of the foci

Answer 51

thats the part i have trouble with

Answer 52

look here to get familiar with what each term represents: http://www.mathwords.com/f/foci_ellipse.htm

and then try to complete the rest yourself. let me know if you get stuck anywhere.

Answer 53

yea i'm not getting the answer

Answer 54

show me all the steps you took to try and get the answer please.

Answer 55

c^2 = 9^2 - 4^2

Answer 56

i get 65 at the end

Answer 57

look at the equation we ended up with:\[\frac{(x-4)^2}{3^2}+\frac{(y+1)^2}{2^2}=1\]and compare it to the standard form:\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]what are the value of \(a\) and \(b\) here?

Answer 58

9 and 4

Answer 59

no - look carefully please

Answer 60

the answer is (4, -1 + or - radical 13)

Answer 61

that is not what I was asking for - look carefully at the two equations above and please tell me what you think are the values for \(a\) and for \(b\).

Answer 62

pay particular attention to the squared symbols

Answer 63

the values you listed before were \(a^2\) and \(b^2\)