**MORNING CLASS:**

We did two questions on ellipses to start things off for the class, here they are:

Now the question didn't ask for a sketch but Mr.K told us to make one anyways.

**However without making the sketch this question would be real tough to answer anyways so the sketch helps =D. You'll find in a lot of these questions you'll need to make a sketch even though the question itself doesn't state that.**

For this question the sketch allows us to find the values necessary to make an equation of the ellipse. Knowing the

*sum of the focal radii is 6 units*we can find the value of

**a***(semi-major axis)*because

*a is half the value of the sum we get that a is 3 units*. Now that we've found the value of

*a*we now have to find the value of

*b (semi-minor axis)*because that's apart of the equation of an ellipse. We first must

*find the value of c in order to apply the Pythagorean Property c² = a² - b² which will allow us to find the value of b.*To find

*c*we just turn to the

*foci provided by the question*. The

*distance between them is 4 units, the value of c is half that which gives us 2 units*. Now we

*apply the Property, 2² = 3² - b², and get b² = 5 so b = √5. Having those values we get the equation written on the slide.*

I won't explain the solution to this question, it seems pretty straight forward already from the slide. However there were a few important things that sprung from this question that I will talk about. It was very interesting how we just dove into finding the radius and centre of this circle. But how do we know it's a circle? Why not an ellipse? So here's some important things to remember when doing questions that don't state what kind of Conic it is:

**- equations of circles always have the same coefficient value on the terms x² and y²**

- equations of ellipses have different coefficient values on the terms x² and y²

- equations of horizontal parabolas have a y² term and a x term

- equations of vertical parabolas have a x² term and a y term

- equations of ellipses have different coefficient values on the terms x² and y²

- equations of horizontal parabolas have a y² term and a x term

- equations of vertical parabolas have a x² term and a y term

**(VERY IMPORTANT REMEMBER!!!)**

For the next part of class Mr. K just went over in detail the

**Focal Radii Property**and

**Pythagorean Property**of the ellipse

**(check slides 3 and 4 for what is potentially going to be dictionary notes!)**

Then we picked up our sheets with the big circle on it (got from the beginning of class) and drew a point just outside of the circumference preferable 2 cm. Then drew 25-30 points on the circumference having making the points closest to the outer point jumbled up together. Then just like the other day when we folded the paper to make an ellipse we did the same kind of folding method to make......A HYPERBOLA! =D Then DING DING! Class ends.

**AFTERNOON CLASS:**

Having done the folding in the morning class, we sketched out the hyperbola we just created by making the folds. Then we drew in the foci points and discovered a few things....

**- line PF1-line PF2 = line QF1-line QF2 = line RF1-line RF2, therefore, any point drawn on the hyperbola when connected to the two foci produces two lines with a constant distance between them**

*- labeled A1 and A2 on the slide are the vertices, the distance between the two, line A1A2 is known as the Transverse Axis**- folding the paper so that the foci are on top of each other we see that when we open it up we get a line that's directly perpendicular to the transverse axis, the end points for this line are labeled B1 and B2, this line is called the Conjugate Axis**- half the transverse axis is the semi-transverse axis and that's labeled a**- half the conjugate axis is the semi-conjugate axis and that's labeled b**- drawing a line from the centre of the hyperbola to F1 we find that the value of this line is the same as the value of a line connecting B1 to A1, therefore a² + b² = c²*

- drawing dotted lines parallel to the conjugate axis going across A1 and A2, then drawing dotted lines parallel to the transverse axis going across B1 and B2 we see a dotted box*- using the dotted box we draw a dotted diagonal line from the top left corner to the bottom right corner anda dotted diagonal line from the top right corner to the bottom left corner**- these diagonal dotted lines represent the asymtotes of the hyperbola*

- slides 5 and 6 show the equations of horizontal and vertical hyperbolas**posibility of a PRE-TEST on FRIDAY and TEST on MONDAY!**

oh and next scribe will be........

**VINCENT!**

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