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Mathematics 68 Online
OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

yes

OpenStudy (asnaseer):

can you write it please?

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

do you know how to do that?

OpenStudy (anonymous):

yes but I think I got the wrong equation

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

4(x^2 - 8 + 16)

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

the first step would be to take out any common factors

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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?

OpenStudy (anonymous):

yes

OpenStudy (asnaseer):

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?

OpenStudy (anonymous):

ok

OpenStudy (asnaseer):

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?

OpenStudy (anonymous):

9(y+1) = 9

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

9

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

have a go at doing this...

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

9(y+1) = -9

OpenStudy (asnaseer):

where did this come from?

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

(Y + 1)^2 = 1

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

y^2 + 2y + 1

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

ok

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

i agreed

OpenStudy (asnaseer):

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?

OpenStudy (anonymous):

yes

OpenStudy (asnaseer):

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:\[\frac{(x-4)^2}{3^2}+\frac{(y+1)^2}{2^2}=1\]

OpenStudy (anonymous):

ok

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

thats the part i have trouble with

OpenStudy (asnaseer):

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.

OpenStudy (anonymous):

yea i'm not getting the answer

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

c^2 = 9^2 - 4^2

OpenStudy (anonymous):

i get 65 at the end

OpenStudy (asnaseer):

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?

OpenStudy (anonymous):

9 and 4

OpenStudy (asnaseer):

no - look carefully please

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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 \(b\).

OpenStudy (asnaseer):

pay particular attention to the squared symbols

OpenStudy (asnaseer):

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

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