Time to say Goodbye

It seems like yesterday we have started our journey.
It was my pleasure having you as my students.
I wish you all the best during tomorrows exam and in your future endeavors.
Take care.
Mr.P

OH NEVERMIND!

I was looking at the Semester 2, Hahaha

To Mr. P

Dear Mr. P,

The links don't work :S

Conics: Parabolas on a Coordinate Plane

On Tuesday, we learned how to make parabolas on a coordinate plane.

Recall what you learned about parabolas last year. Remember the standard rules of a parabolas shape where if |a| = 1, then the shape would be normal but if |a| > 1 then the parabola would get narrower and if |a| < 1 then the parabola would get wider.

Recall the standard form of parabolas.

(y-k) = a(x-h)^2 or
(x-h) = a(y-k)

For our Conics unit, we have learned the General Form for a Parabola:

Ax^2 + Cx + Dy + E = 0 or
By^2 + Cx + Dy + E = 0

We use completion of squares to convert to standard form to find the vertex.

Example 1: Find the vertex, direction and shape of parabola, axis of symmetry, domain and range for each conic.

2x^2 + 4x - 4y +22 = 0

We can't find any data with this kind of equation so ee need to complete the square in order to solve this equation.

2x^2 + 4x - 4y + 22 = 0
2(x^2 + 2x) - 4y = -22
2(x^2 + 2x + 1) - 4y = -22 + 2
(2(x + 1)^2 - 4y = -20) / 2
(x + 1)^2 = 2y = -10

Rearrange the equation...

-2y = -(x+1)^2 - 10

Now isolate the y...

(-2y = -(x+1)^2 - 10) / -2
y = (1/2)(x + 1)^2 + 5

Now we can fill in the missing data with this equation.

Vertex: (-1, 5)
Direction: Upwards
Shape: Wide
Axis of Symmetry: x = -1
Domain: (-∞, ∞)
Range: (-5, ∞)

Example 2: If the parabola y^2 + x - 6y - m = 0 passes through the point (-3,5) find the vertex.

First we need to solve for m. We do this by plugging in the values of x and y.

(5)^2 + (-3) - 6(5) - m = 0
25 - 3 - 30 = m
m = -8

Now we can solve the equation.

y^2 + x - 6y - (-8) = 0
y^2 - 6y + x + 8 = 0
(y^2 - 6y + 9) +x + 8 - 9 = 0
(y - 3)^2 + x - 1 = 0
(x-1) = -(y-3)^2

Therefore, the vertex is at (1,3).

QED

Also a reminder that all assignments are due in Wednesday. Please submit any work that you have by then!

Good luck! :D

Conics - Circles on a coordinate plane

Hey everyone

Today we started our conics unit, and the first topic is circles on a coordinate plane. There are a few important equations to remember:

Standard form of a circle: x2 + y2 = r2 AND (x - h)2 + (y - k)2 = r2

General form of a circle: Ax2 + By2 + Cx + Dx + E = 0

Area of a circle: π r2

Circumference of a circle: 2 π r

Distance between two points: D = √(x2 - x1)2 + (y2 – y1)2

Midpoint between two points: MP = (x2 + x1) , (y2 + y1)
2 2


The most basic operation with a circle is to find the center and radius.
Ex.:
(x-3)2 + (y+4)2 = 48

we find the center by using the h and k values, and reading them as opposites.

Center @ (3, -4)

the radius is the square root of the number after the equal sign.

radius is √48


The second operation we worked on is finding the center, radius, domain, and range of a circle in general form.
Ex.
x2 + y2 – 6x – 10y – 2 = 0

We reorganize the equation so that all like terms are together.

x2 - 6x + y2 – 10y = 2

Then we complete the squares by dividing the C and D values by 2 and squaring them. We add these numbers to both sides.

x2 - 6x + 9 + y2 – 10y + 25 = 2 + 9 + 25

Then we solve each set for one binomial that is squared. Now it is solved.

(x - 3)2 + (y - 5)2 = 36

Next, we find the canter and radius the same way as before.

Center @ (3, 5)
Radius = 6

After sketching the circle, the points are at (-3, 5), (3, -1), (3, 11), and (9, 5). This helps us find the domain and range.

Domain = [-3, 9]
Range = [-1, 11]


The next operation is very similar, but has coefficients.
Ex.
16x2 + 16y2 – 8x – 143 = 0

16x2 – 8x + 16y2 = 143

16(x2 – 1/2x) + 16y2 = 143

We do the same thing with dividing the second number by two, and squaring it.

16(x2 – 1/2x + 1/16) + 16y2 = 143

Then we divide all the numbers in the equation by 16.

16(x2 – 1/2x + 1/16) + 16y2 = 144
16

We then continue to solve as normal.

(x – 1/4)2 + y2 = 9

center @ (¼, 0)
radius = 3


The last operation is using the distance and midpoint formulas.
Ex.
End points at (-2, 8) and (3, 4).

D = √(x2 - x1)2 + (y2 – y1)2
D = √(-2-3)2 + (8-4) 2
D = √(-5)2 + (4) 2
D = √25 + 16
D = √41

Radius = √41/2

MP = (x2 + x1) , (y2 + y1)
2 2

MP = (3 + 2) , (4 + 8)
2 2

MP = (½, 12/2)

MP = (½, 6)

Center @ (½, 6)

Then we plug it in to the formula:

(x – 1/2)2 + (y - 6)2 = √41/2

Happy Name Day Mr. Piatek !!

I bid you a joyful name day and wish you a prosperous future with many more name days to follow!

November Hall of Famer

And the winners are:

1 - Vince 5 points - 3 bonus points
2 - Kendra, Sharlyne, and Mark - 3 points - 2 bonus points each