What are the foci of the ellipse given by the equation 4x^2 + 9y^2 – 32x + 18y + 37 = 0?
do you know what the standard form for the equation of an ellipse is?
yes
can you write it please?
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
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
so, what you need to do is to first convert your equation into this form
do you know how to do that?
yes but I think I got the wrong equation
if you write out your steps then I can help you spot where you may have gone wrong
i completed the squares and all that and I end up with the equation (x-4)^2/93 + 3(y+1)^2/124
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
it is usually simpler to concentrate on just the x components, and then just the y components
so lets start with:\[4x^2-32x\]how can you complete the square here?
4(x^2 - 8 + 16)
\[4x^2-32x\ne4(x^2-8+16)\]
the first step would be to take out any common factors
so we start with:\[4x^2-32x=4(x^2-8x)\]
I meant - take out any constant factors (i.e. numeric values only)
now try and complete the square on:\[x^2-8x\]
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?
yes
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?
ok
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?
9(y+1) = 9
thats not correct - look carefully at how we worked with the x terms and try again please.
what is the common constant factor in:\[9y^2+18y\]
9
correct, so we can first simplify this to:\[9y^2+18y=9(y^2+2y)\]
next we try and complete the square for:\[y^2+2y\]
have a go at doing this...
dont u have Y^2 + 2y + (2/2)^2
y^2 + 2y + 1 y^2 + 2y = -1 9(y+1) = -1
I don't understand where you are getting the last line from?
9(y+1) = -9
where did this come from?
we are trying to complete the square for:\[y^2+2y\]
(Y + 1)^2 = 1
ok - what would you get if you expanded:\[(y+1)^2\]
y^2 + 2y + 1
correct, so we can therefore say that:\[y^2+2y=(y^2+2y+1)-1=(y+1)^2-1\]
ok
so, lets summarise what we have found so far...
\[4x^2-32x=4(x-4)^2-64\]\[9y^2+18y=9(y^2+2y)=9((y+1)^2-1)=9(y+1)^2-9\]agreed?
i agreed
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?
yes
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:\[\frac{(x-4)^2}{3^2}+\frac{(y+1)^2}{2^2}=1\]
ok
you should now be able to find the coordinates of the foci
thats the part i have trouble with
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.
yea i'm not getting the answer
show me all the steps you took to try and get the answer please.
c^2 = 9^2 - 4^2
i get 65 at the end
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?
9 and 4
no - look carefully please
the answer is (4, -1 + or - radical 13)
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 \(b\).
pay particular attention to the squared symbols
the values you listed before were \(a^2\) and \(b^2\)
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