Conics: Ellipses:
    Finding Information from the Equation (page 2 of 4)

Sections: Introduction, Finding information from the equation, Finding the equation from information, Word Problems

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 16x² + 25y² = 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 x² = (x – 0)² and y² = (y – 0)², the equation above is really:

Then the center is at (h, k) = (0, 0). I know that the a² is always the larger denominator (and b² 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 a² = 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 a² and b², so:

a² – c² = b²
25 – c² = 16
9 = c² Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved

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
  25x² + 4y² + 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.

25x² + 4y² + 100x – 40y = –100
25x² + 100x + 4y² – 40y = –100
25(x² + 4x       ) + 4(y² – 10y        ) = –100 + 25(    ) + 4(     )
25(x² + 4x + 4) + 4(y² – 10y + 25) = –100 + 25( 4 ) + 4( 25 )
25(x + 2)² + 4(y – 5)² = –100 + 100 + 100 = 100


The larger demoninator is a², 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, a² = 25 and b² = 4, so the equation b² + c² = a² gives me 4 + c² = 25, and c² 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

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