MORNING CLASS
We started off the morning with word problems.
1.
This problem was solved by yours truly (ahem haha). From the given question, we are given that the foci are (+6,0) and the focal radii is 4 units. Since we know what the focal radii is, we could determine the vertices of the hyperbola by subtracting the focal radii from the foci. Doing those, we should get (+2,0) as our vertices. Thinking about it, the x-value of the vertices are the same as the a value in an equation of a hyperbola. We then get that the a value is 2. We already know the c value of the ellipse equation, so that's enough information to solve for b by using the Pythagorean identity, c2=a2+b2. Substituting are values, we then find out that the value of b is the square root of 32, or 4 rt.(2). Now that we have our respective values, we can now write our equation.
In this question, it doesn't matter what form you leave your equation on because it doesn't specify whether you leave your equation in standard equation (the equation above) or in general form (the equation below). Moving on...
2.
This problem should be pretty straightforward as it is like one of the problems we did yesterday/2 days ago.
3.
This problem is a bit different from the ones we did. In this problem we were given the directrix of the parabola (x-2=0 or x=2) and the vertex (-1, 3). From that we can determine that the parabola opens to the left. Also, from the data given, we can also determine the p value, which is 3. Now that we have enough data, we can determine the equation, which is on the image, in red colour.
Note: the equation underlined in blue is wrong because we forgot to consider the vertex.
4.
Now on to the "real" word problem. This problem requires some thought and some manipulation. From the graph given, we can see that we drew a Cartesian plane on the graph so it would make it a bit easier for us. We plotted out several points on the graph like (10,0), which is the distance from the origin/cliff, (0,40), which is the vertex of the graph/height of the cliff and (?,30) which is the point that we're trying to find. From the solution on the image, we can see that we used the point (10,0) to determine the value of p, which is -5/8. Now that we've found p, it's as easy as substituting the values and solving for x. We then find out that x = +5, but we reject -5 because if we throw -5 distance, we're throwing into the cliff. Also Mr. K pointed out that since it's a word problem, we ALWAYS have to put our answers in a sentence.
5.
This problem seems to be straightforward, but requires a bit of thinking, like any other else. We know that the center is at (0,0) as it is given, and a vertex which is (rt.(6), 0). Thinking back, we know that the x-value of the vertex is equal to the a value in the equation of an ellipse. Now that we know a, we can solve for b by putting in 9 for the x value and 5 for the y value. We then find out that b = rt.(2).
We end our morning class here.
AFTERNOON CLASS
We started our afternoon class with a quiz. I'm kinda hesitant on putting up the solutions here because some of us haven't written the quiz. But if you want to see it, check today's slides. After the quiz, we went to our groups and solved a couple of problems.
1.
This question got most of us wondering. But Grey-M figured it out and once he put up the solution, we were all like, "oooohhh I see I see". It seems pretty straightforward.
2.
We didn't really get to finish this problem because the bell rang, beckoning us to go home and have fun!! But we did the first part of the problem, which seemed pretty straightforward.
The next scribe will be Aichelle.
NOTE: THE PRE-TEST IS ON MONDAY MORNING, THE TEST IS ON MONDAY AFTERNOON. WE WERE SUPPOSED TO HAVE THE PRE-TEST TOMORROW BUT WE HAVE BUS RIDERSHIP TOMORROW SO MR. K MOVED IT ON MONDAY MORNING. TOMORROW'S CLASS WILL BE A WORKSHOP.
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