Module 7: Conics
Lesson 2 Notes
Ellipse
Video 2 Transcript
Hey everyone this is a video for Module 7 lesson 2 ellipse notes video 2.
In this video we are going to take this ellipse and find the following information. So
(𝑥+2)2
(𝑦−5)2
we have 4 + 49 = 1. So the first thing we are going to find is the center of
this ellipse. The center of this ellipse is going to be (-2, 5). So I am going to go to the
next page and I am going to go ahead and start graphing this. At -2, 1, 2, 3, 4, 5, that
is going to be our center. If we analyze this real quick, we have “a” squared equals 4
and “b” squared equals 49. That means that the “a” value is 2, and the “b” value is 7.
“a” is over your “x” squared term so on your, x, running left to right since the x-axis
runs from left to right you are going to go right 2 units and left 2 units, from the
center. So you can actually do that by just doing addition there. You center is (-2, 5)
if I am adding 2 to the right, that is actually adding 2 to the x-coordinate and leaving
“y” alone, and here you are taking -2 and subtracting 2 and leaving the “y”
coordinate alone. So this is going to become (0,5) and this will become (-4,5). So we
have actually found in those 2 points and if you remember in your notes those were
important. So I am going to just go ahead and label those (0,5) and (-4, 5). So once
again, I got those by adding the “a” value to the “x” coordinate.
Ok, and “b” equals 7. I am going to go to the center and rise up 7and go down 7. So
from the center, I go up 7 and from the center I go down 7. Once again I can do that
without a graph. If you take your center and know that the y-value I am going to go
up 7, so I would leave the -2 as is and add 7 to that, and I would also subtract 7 from
that. So I would get (-2,12) and (-2, -2). So this would be (-2, -2) and this one up
here would be (-2,-12). I am going to go ahead and attempt an ellipse there. So
there is my ellipse.
Now as you can visually see this is going to be my major axis if you remember from
the notes, the major axis is going to be the longer axis going through the center. So I
can either have this one, or the horizontal one. The vertical one is definitely the
longer one here. You can also verify that just by looking at the number below each
coordinate. So the y-axis runs vertically this number is larger than this number with
the “x’ which runs horizontally. So with “b” squared be greater than “a” squared, my
major axis is going to be this way. So my major axis is going to run vertical and its
going to be length of 14 units because I moved 7, I went up 7 and down 7 so that is a
total of 14 units. And my vertices are the vertices at the end of the major axes is
going to be (-2,12) and (-2, -2).
My minor axis would be the smaller axis of the 2. Its going to run horizontal, its
going to be 4 units long because my “a” was 2, double that and it would be 4 units.
And my covertices which are the end of my minor axes, are (0,5) and (-4, 5).
Ok, so lets go back and see what we have found. We have found the vertices the
covertices, the length and direction of the major and minor axes, and now we have
to find the coordinates of the foci. So the coordinates of the foci are going to fall on
the major axis. So I need this relationship in order to find the foci. Remember in
your notes I related those 2 by, if the “b” squared is greater than the “a” squared, I
do “b” squared minus “a” squared equals “c” squared. “c” is going to be the value of
the length of the foci, which is measured from the center out towards the ellipse on
the major axes. So looking back, our “b” squared was 49 and our “a” squared was 4.
So I have 49-4 equals “c” squared, which is 45. And “c” is going to be the square root
of 45. To get an approximate answer there, I am going to go ahead and do the square
root of 45 in my calculator. Which is going to be approximately 6.708. So what
happens here I take this value and add it to my center, so my center is (-2,5) so since
my major axis runs vertically I am going to take this amount and add it to my ycoordinate. So to get the coordinate of the foci, I am going to take -2 and I am going
to add 6.708 and I am going to take -2, and subtract 6.708 from 5. So in both of
these cases I am adding it to the y-coordinate. So 5 plus 6.708 and 5 minus 6.708.
So to find those actual coordinates I am going to do the math there, so, and those
points would be (-2, -1.708) and (-2,11.708). ]
11.708 is going to be right here, and 1.-708 is going to be around here (blue dash
marks). So the foci coordinates graphically are here and here. The coordinates are
these 2.
And the last thing we have to find is the eccentricity and you have to compare the “c”
value with either the “a” or the “b” value which ever one is bigger. So we have said
several times that the “b” is the biggest here, the largest. So we take the “c” value and
divide by the “b” value would equal the eccentricity. The eccentricity as you read in
your notes measures, how oval the ellipse is. So I am going to take the “c” value
which was the square root of 45 and divide that by the “b” value, which was a value
of 7.
Remember reading in your notes the close to 1 the more oval it is, the closer to 0 the
more circular it is. So the eccentricity of this is going to be .958, so that is pretty
close to 1. So you can see the ellipse is pretty oval. So that is how you find all the
pieces of that ellipse