Week 3: Rotation of Conics

We looked at the rotation of conics. In an equation it would have Ax^2 +Bxy +cy^2+ Dx + Ey + F=0

There cannot be a Bxy in an equaiton at all.

To get rid of it you first

Step 1: Find the angle with cot2θ=A-C /B 0<θ<90

Step 2: You replace x&y x=x^1 cosθ-y^1 sinθ and y=x^1 sinθ+y^1cosθ

Step 3: Then you use Algebra to simplify

It sounds easy at first but it is so much more.

Here’s some helpful hints:

1+cot^2 θ=csc^2 θ

cos2θ

sin2θ

and

sinθ=√1-cos2θ/2

cosθ=√1+cos2θ/2

Here is one problem below

It is a lot more complicated. Just like the steps you plug in to find the angle and then replace the x and y to find the answer and then simplify. 

Below is another example:

This time, instead of looking for the angle, you put it into the sinθ/cosθ equation and solve. 

Then there will be times when you solve but you just want to find out what shape the equation is. Then you use 

B^2 - 4AC = 0 Parabola

B^2 - 4AC < 0 Ellipse

B^2 - 4AC > 0 Hyperbola

Below is an example and then two problems

This is how you solve rotation of conics

SEE YOU NEXT WEEK!!!

Week 2: Parabolas

This Week we looked at Parabolas. Parabolas are U shapes on a graph. 

To find a parabola you use: (x-h)^2=4c(y-k)

vertex: (h, k)

focus: (h, k+c)

directrix=y=k-c

axis of symmetry: x=h

The focus is a point above the vertex of the equation and it is always the C away from the vertex.

The directrix is below the vertex and it is a line that runs parallel to the vertex and is C away from the vertex. 

The above equation is for a vertical equation.

A Horizontal equation would use:

(y-k)^2 =4c(x-h)

vertex= (h,k)

focus: (h+c, k)

directrix: (x=h-c

axis of symmetry: y=k

Below is an example problem on the parabola. 

This is how you solve parabolas and graph them

SEE YOU NEXT WEEK!!!