Today we had the joy of having a double class today. A lot of lectures and writing and diagrams. That was all we had done today the people in our class could give you what we did =P.
Just joking, the first class started good with two questions on the parabola's; basically a review from yesturday.
For each parabola we were supposed to find
a) sketch the graph
d) Equation of directrix
e) Axis of Symmetry
- 20x + 2y + 1 = 0
First thing's first, we figure out which way it opens, and seeing as y is being squared we notice that it opens either right or left. Therefore we isolate all the 'y' onto one side and everything else on the other.
+ 2y = 20x - 1
Secondly we have to complete the square for the 'y' in order for us to get that nice ( y - k )2
+ 2y + 1 = 20x -1 + 1
Then we just simply simply and factor out.
( y + 1 )2
= 20 x
With that figured out, we now had to find 'p'. To find p we simply set 4p = 20 because nomrally where 20 would be 4p would be, and we get :
4p = 20x
p = 5
Simply just divide both sides by four in order to isolate 'p.'
With all the information we figured out we could basically figure out the questions that were asked.
b) Vertex (0, -1)
c) Focus ( 5, -1)
d) Directrix: x = -5
e) Axis of Symmetry: y = 2
The sketch should be easily figured out now with the answers. I will provide a sketch for you guys if you would like a reference, or if you're having trouble sketching it. Try it first, because me telling you and showing won't get you anywhere.
- 4x + 8y + 4 = 0
This question was done exactly the same was as the first question. The only difference here is now notice that the parabola opens up or down therefor we isolate the x's on one side and the other's on the other. I'm pretty sure you wouldn't like me to show you again how this is all done but unless you want me too i'll go back and edit =P
Still in the first period...
Remember that paper we manipulated, abused and everything else you could do with a peice of paper ? Well today we put it into good use. We created an ellipse with the folds that we had made, similiar to the parabola that we manipulated with the other peice of paper. With the Ellipse we made 3 different points on the ellipse and constructed lines to connect them to the two dots we had made, which we found out was called 2 focus' or foci? (spelling ?). From this construction we found a pattern to show that any line from a point on the perimeter of an ellipse to the focii had all the same sums. The lines we constructed were also called 2 focal radii. We then constructed the major axis, which is the longest line in the ellipse that crosses the center and found that it was also the same length of two focal radii. Afterwards we then created the minor axis, which is the shortest line from one point on the ellipse to another point on the ellipse passing the center. Using these two axis' we constructed a right angle triange by connecting on of the points of the minor axis to one of the points. Btw we also found out that a semi minor or major axis is half of the major or minor axis. We then used the pythagorean thearom to figure out the hypotenous of the triange.NOTE : THIS IS NOT LIKE THE THE ONE WE LEARNED IN JUNIOR HIGH IT IS NOT A2 + B2 = C2
The one we had learned is a2
. This is because since the that created the hypotenous is equal to the semi major axis we defined that as a and major axis A. The semi minor axis labeled b and the minor axis labelled B. C was the distance from the centre of the ellipse to one focus.
With that we had learned everything we needed to know about ellipses, and went straight into our dictionaries.
Definition : the locus of a point that moves in such a way so that the sum of it's distances (focal radii) from two fixed points (the foci) is constant
The Anatomy of an Ellipse
A more simplified diagram of our folding method but less confusing and more neat.
We learned about the standard fomula of an ellips
( x - h )2
+ ( y - k )2
( x - h )2
+ ( y - k )2
The joy of copy and paste =D, haha anyway we learned that the difference of the two equations was in fact a and b. This was because since a is the bigger line which ever it is the denominator for (either x or y) depends where it will stretch either up and down or left and right. Also since b is the smaller side, similiarily like a, which ever it is the denominator for (either x or y) depends whether the ellipse will either become compressed either up and down or left or right.
We also found out where c was. from the formula a2
we notice that the denominators are in fact a and b in the formula. There fore in order to find c. We place in the values of a and b and solve for c. I'm pretty sure a and be will already be given in their squared form so you do not have to worry about squaring them. I am not 100% sure about this but Mr. K did not say and no one asked and I just questioned it now. SO MR. K IF YOU ARE READING THIS (which i know you are) I WOULD LIKE TO KNOW IF IT IS ALWAYS SQUARED ?
From here we did a couple problems on sketching and everything else like we did in the first period but this time working with the standard formula. It was basically the samething you have the vertex at (h,k). Then solving for c using the 'new' pythagorean thearom. From there you could sketch the graphs and that's all about ellipses.
If a question is given in general form you would do the samething as you would do for a circle. Isolate the y2
and the x2
on to one side and the constant on to the other. Factor out anything that could be factored, complete the square, then to make everything equal to 1, you divide that one side by itself to make it one and everything else so that you do not get a constant with the x's and y's. and that's it the whole bottle of wax.
btw the next scribe is zaenab.
homework PG 150 1 - 27 ODD ALSO 32 - 38