Your first task will usually be to demonstrate
that you can extract information about an ellipse from its equation, and
also to graph a few ellipses.

State the center, vertices, foci
and eccentricity of the ellipse with general equation 16x^{2}
+ 25y^{2} = 400,
and sketch the ellipse.

To be able to read any information from
this equation, I'll need to rearrange it to get "=1",
so I'll divide through by 400.
This gives me:

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Since x^{2}
= (x – 0)^{2} and
y^{2} =
(y – 0)^{2}, the equation
above is really:

Then the center is at (h,
k) = (0, 0). I know that the
a^{2} is
always the larger denominator (and b^{2}
is the smaller denominator), and this larger denominator is under the
variable that parallels the longer direction of the ellipse. Since 25
is larger than 16,
then a^{2} =
25, a =
5, and this ellipse is wider (paralleling
the x-axis)
than it is tall. The value of a also
tells me that the vertices are five units to either side of the center,
at (–5, 0)
and (5, 0).

To find the foci, I need to find the
value of c.
From the equation, I already have a^{2} and
b^{2},
so:

Then the value of c is
3,
and the foci are three units to either side of the center, at (–3,
0) and (3,
0). Also, the value of the eccentricity
e
is c/a =
3/5.

To sketch the ellipse, I first
draw the dots for the center and the endpoints of each axis:

Then I rough in a curvy line, rotating
my paper as I go and eye-balling my curve for smoothness...

...and then I draw my "answer"
as a heavier solid line.

center (0,
0), vertices
(–5,
0) and (5,
0), foci (–3,
0) and (3,
0),
and eccentricity 3/5

You may find it helpful to do the roughing
in with pencil, rotating the paper as you go around, and then draw your
final graph in pen, carefully erasing your "rough draft" before
you hand in your work. And always make sure your graph is neat and is
large enough to be clear.

State the center, foci,
vertices, and co-vertices of the ellipse with equation 25x^{2}
+ 4y^{2} + 100x
– 40y + 100 = 0.
Also state the lengths of the two axes.

I first have to rearrange this equation
into conics form by completing the square
and dividing through to get "=1".
Once I've done that, I can read off the information I need from the
equation.

The larger demoninator is a^{2},
and the y part
of the equation has the larger denominator, so this ellipse will be
taller than wide (to parallel the y-axis).
Also, a^{2}
= 25 and b^{2}
= 4, so the equation b^{2}
+ c^{2} = a^{2}
gives me 4 + c^{2}
= 25, and c^{2} must
equal 21.
The center is clearly at the point (h,
k) = (–2, 5). The vertices
are a = 5
units above and below the center, at (–2,
0) and (–2,
10). The co-vertices are b
= 2 units to either side of the center,
at (–4, 5)
and (0, 5).
The major axis has length 2a
= 10, and the minor axis has length
2b = 4.
The foci are messy: they're sqrt[21]
units above and below the center.

center (–2,
5), vertices
(–2,
0) and (–2,
10), co-vertices
(–4,
5) and (0,
5), foci
and , major
axis length 10,
minor axis length 4

Stapel, Elizabeth.
"Conics: Ellipses: Finding Information from the Equation."
Purplemath. Available from http://www.purplemath.com/modules/ellipse2.htm.
Accessed