Expert: Steve Holleran - 8/9/2008

Question
QUESTION: find an equation of the ellipse having a major axis of length 12 and foci at (5,9) and (5,-1)

ANSWER: Hi John,

Okay, from what I recall about ellipses , the general equation of a vertical ellipse is

  (x - h)^2 / b^2 + (y - k)^2 / a^2 = 1

where (h, k) are the coordinates of the center, and 2a = length of major axis and 2b = length of minor axis.

Now, the foci always lie on the major axis, so if one focus is (5,9) and the other (5, -1), then the ellipse is oriented vertically oblong.

The distance between the foci is 10, so from the center to each focus is 5.  That makes the center here at (5, 4).

Also, you have 2a = 12, so a = 6, and we know the distance from the  focus to the center to be 5, and this is called c , so c = 5.

Now, in a vertical ellipse, the a,b,c variables satisfy the equation

         a^2 = b^2 + c^2  so we have 6^2 = b^2 + 5^2

or          36 = b^2 + 25  so b^2 = 11.

That makes the equation

         (x - 5)^2 / 11  + (y - 4)^2 / 36 = 1


Hope this is what you needed.
Steve

---------- FOLLOW-UP ----------

QUESTION: put the hyperbola -4x^2 + 9y^2 -32x +90y +17 = 0  in standard form

Answer
Hi John,

You want to group the x and y terms together and complete the square:

   -4x^2 - 32x  + 9y^2  + 90y = -17

   -4(x^2 + 8x     )  + 9(y^2 + 10y     ) = -17

Now to complete the square in x, you need +16 in the parenthesis, but notice that it would be multiplied by -4 in the front, so you're actually adding -64.  So do that on both sides, and for the y's, you need to add +25, which is actually 9*25 = 225:

   -4(x^2 + 8x + 16) + 9(y^2 + 10y + 25) = -17 -64 + 225

   -4(x+4)^2 + 9(y + 5)^2 = 144

Now divide all terms by 144:

     -(x+4)^2 / 36 + (y+5)^2 / 16 = 1

or    (y + 5)^2 / 16 - (x + 4)^2 / 36 = 1

That's your standard form

Steve