December 14, 2006

Word Problems with Conics

Hey kids, I was yesterday's scribe post. This is going up a day late because I had to work late last night and I went straight to bed as soon as I got home. Anyways yesterday was a one period class, and we worked on some conic word problems from previous exams. Here are the five questions given to us on the board:

1. Find the equation of an ellipse with major axis AB and minor axis CD if the following coordinates are the vertices: A(2,7) B(2,-1) C(0,3) D(4,3).

2. Identify each conic:

3. Find the equation of the circle with the centre (4,0) which passes through the orgin.

4. y²-2y-3=0

a) Find all intercepts.
b) Sketch a clearly labeled graph.

5. A balloon arch in the shape of one branch of a hyperbola is made for a wedding party. The arch is to span 4m and have a max clearance of 2.2m.

a) Find an equation for the hyperbola.
b) What would be the height of the arch 0.8m from one end?

ANSWERS:





1. This is what I know. AB have the same x-coordinates, so that means that the line of AB runs vertically. So I just added the absolute values of the y-coordinates to get the length of AB: -1 + 7 = 8. It's an absolute value because distances don't have negative values. So AB is 8 units long, and we divide it by 2 to get the value of a, which is 4. So, we can say that line of CD runs horizontally and we add together their x-coordinates: 0 + 4 = 4. We divide 4 by 2 to get the value of b and we get 2.





To get the centre, we take the y-coordinate of A and subtract the value of a from the y-coordinates, just as we take the x-coordinate of D and subtract the value of b from the x-coordinates.

If you wanted to add the value of a to find the y-coordinate of the centre, you'd have to add from B's y-coordinate. To add the value of b to find the x-coordinate of the centre you'd have to add it from C's x-coordinate.

Subtracting

AB: 7 - 4 = 3
CD: 4 - 2 = 2

Adding:

AB: -1 + 4 = 3
CD: 0 + 2 = 2

So the center is (2,3).

Now we can start plugging in our values to an equation.



2. a) We can say that it is a hyperbola, because we have the subtraction sign between a fraction of x² and a fraction of y².

b) In the expansion we have positive squared values for x and y, which tells us it can't be a parabola or a hyperbola, but they have different coefficients so that tells us it is has to be an ellipse.

3. The equation of a circle is: (x-h)² + (y-k)² = r²

(h,k): (4,0)
point: (0,0)

From this we can tell that both points rest on the x-axis, and have are 4 units away from each other. Hence, we can say that the radius of this circle is 4 units.

(x-4)² + y² = 4²
(x-4)² + y² = 16

4. We take our eqation and change it to standard form.

y² - x - 2y - 3 = 0
(y² - 2y + 1 ) = x + 3 + 1
(y-1)² = x + 4


a) To find intercepts we just plug in zero for our variable.

x--intercept; y = 0

(0 - 1)² = x + 4
(-1)² = x + 4
1 = x + 4
-3 = x

y-intercepts; x = 0

(y - 1)² = 0 + 4
(y - 1)² = 4
y - 1 = ±2 [square root both sides]

y -1 = 2
y = 3

y - 1 = -2
y = -1

Therfore, our x-intercept is at -3, and our y-intercepts are at 3 and -1.

b)
















5.

After realizing how lengthy question 5 was, we got to doing some group work. I kind of lost the sheet with the question on it, so I don't have much else to post up. I'll fix this up when I do find it, I just kind of forgot I never finished this. Sorry for the huge delay!

1 comment:

  1. where's the answer to number 5

    ReplyDelete