Another Focus of an Ellipse

HCL

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What are the foci of the ellipse

16( y - 3)^2= 16 - (x - 2)^2 ?




These problems are killing me. Test in the morn, still don't get it. First, I divided by 16 to get (y-3)^2 = -(x-2)^2/16
and then convert that to

(y-3)^2 + (x-2)^2/16 = 0 factor out to y2-6y+9 + (x^2-2x+4)/16=0, put the 9 on other side and divide out to get...

y^2/9 - 2y/3 + (x^2-2x+4)/144=1 , then i get a common denominator by multiplying by 3 to get (y^2-6y)/9 +(x^2-2x+4)/144=1

and here i'm stuck... i 'm not sure if i'm making a mistake in the above or not seeing the next step to find a, b
 
Your algebra needs some work. You can't work on conics with poor algebra skills. Maybe you're just stressed and forgetting.

16( y - 3)^2= 16 - (x - 2)^2

A little closer to Standard Form

(x - 2)^2 + 16( y - 3)^2 = 16

Now divide by 16

(x2)216+(y3)2=1\displaystyle \dfrac{(x - 2)^2}{16} + ( y - 3)^2 = 1

Notice how 16/16 = 1. Somehow, you managed zero in your demonstration.

Does it look more clear, now?
 
This is a shifted conic, centered at (h,k).

The foci are located at (h-c, k) and (h+c, k).

Check out this page. Scroll down to the paragraph that begins with "The graph of an ellipse can be translated so that its center is at the point (h, k)".

Earlier on that page, they provide the relation c^2 = a^2 - b^2.

Cheers :cool:
 
Ok so I made it thru the webpage etc. But do I need to worry about the -2 and -3 or can 16=a and 1=b so c^2 = the sqr root of 15? Plus or minus.



Your algebra needs some work. You can't work on conics with poor algebra skills. Maybe you're just stressed and forgetting.

16( y - 3)^2= 16 - (x - 2)^2

A little closer to Standard Form

(x - 2)^2 + 16( y - 3)^2 = 16

Now divide by 16

(x2)216+(y3)2=1\displaystyle \dfrac{(x - 2)^2}{16} + ( y - 3)^2 = 1

Notice how 16/16 = 1. Somehow, you managed zero in your demonstration.

Does it look more clear, now?
 
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Ok so I made it thru the webpage etc. But do I need to worry about the -2 and -3 or can 16=a and 1=b so c^2 = the sqr root of 15? Plus or minus.
The -2 and -3 tell you where the CENTER of the ellipse is.
........O = (2,3)
The foci are in the ±x direction measured from point O, at distance sqrt(15).
 
do I need to worry about the -2 and -3

Yes! They tell you the values of h and k. You need to know these values, to locate the foci.

(x-2)^2 is of the form (x-h)^2

From this, you ought to "see" that h equals 2.

Likewise, (y-3)^2 is of the form (y-k)^2, so k must be 3.

You already found ±c correctly.

The formula for the foci coordinates uses h, k, and ±c. It adds c and -c to the x-coordinate of the center (that's h):

(h+c, k)

(h-c, k)

You know the values for h, k, and c. Substitute and do the arithmetic :cool:
 
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