Expert: Paul Klarreich - 7/15/2007

Question
(conics)"Find the foci of an ellipse with the equation: 7(x-2)^2+3(y-2)^2=21.  
I
think it might be (4,2) and (0,2) but that's only a guess.  I just don't understand
what steps to do.  Don't I set something to zero?  Help please!!

Answer
Questioner:   Alexandria
Category:  Calculus
Question:  (conics)"Find the foci of an ellipse with the equation: 7(x-2)^2+3(y-2)^2=21.  
I think it might be (4,2) and (0,2) but that's only a guess.  I just don't understand what steps to do.  Don't I set something to zero?  Help please!!

..............................
Hi, Alexandria,

(Mostly I'm reviewing this for my own benefit, to see if I remember the stuff.)

The standard form for an ellipse is:

WARNING: VIEW THIS IN A FIXED SIZE FONT, LIKE COURIER.

x^2   y^2
--- + --- = 1
a^2   b^2

or

x^2   y^2
--- + --- = 1
b^2   a^2

where we always  have  a > b.  

As I recall, the definition of an ellipse is:

The set of all points such that the sum of their distances to two fixed points (called the foci) is a constant.

Now imagine that our ellipse has:
1. Center at (0,0)
2. Foci along the x-axis, at F1(c,0) and F2(-c,0) so the x-direction is longer. (A HORIZONTAL ellipse - the foci are right and left of the center.)

Then a point at the right end of the ellipse, at  (a,0) will have distances  a-c to F1 and a+c to F2, thus a sum of  2a.  That's our constant.

Further, if we take a point (0,b) on the y-axis, its distance to F1(c,0) must be 'a', and the Pythagorean theorem gives:

a^2 = b^2 + c^2  or   c^2 = a^2 - b^2,  or   b^2 = a^2 - c^2

That's the rule that defines the relation between a,b,c:  a is biggest.

Also, if the foci are along the y-axis, at  F1(0,c) and F2(0,-c), this is a VERTICAL ellipse, and the foci are above and below the center.  

Further, if the ellipse looks like:

(x - h)^2   (y - k)^2
--------- + --------- = 1
  ??          ??
then we are offset by (h,k) -- that is the actual center and the foci are +-c either right-and-left or  above-and-below the center.
.........................

Now for your example:

7(x-2)^2+3(y-2)^2=21

To get it into a standard form, the RHS must be equal to 1.  So divide out that 21:
(x - 2)^2   (y - 2)^2
--------- + --------- = 1
   3          7


Now we can match things up and draw some conclusions:

Since  7 is bigger than 3,  a^2 = 7  and  b^2 = 3.
The relation  c^2 = a^2 - b^2 gives:

c^2 = 7 - 3 = 4
c = 2.

So:
1. This ellipse is VERTICAL; the foci are above and below the center.
2. The foci are 2 units above and below the center.
3. The center is at  (2,2)
4. A point 2 units above that is (2,4). That is F1.
4. A point 2 units below that is (2,0). That is F2.