1. The document discusses Luis' notes from a class on conic sections. It provides the problems and solutions from a pre-test on hyperbolas, ellipses, parabolas, and their transformations. 2. Later problems involve finding the equation of an ellipse given its axis endpoints and a point that lies on it, as well as deriving the equation of a parabola given its vertex and directrix. 3. Luis concludes by announcing that tomorrow is the unit test and assigning the next class scribe.

1. Morning Class...
1

2. Greetings! Hello, It's me Luis posting again for
the third time! These are the problems on our
Pre­test on Conics...We start off solving...
(y+3)2 = ­ 1
2
The hyperbola (x­2) ­
1.
4
It was just simply asking it's vertical transverse
axis ... and the answer is 4.
Piece of cake! ryt?
2

3. 2. The image of the conic section 2x2 + 3y2 ­12 = 0
is expanded horizontally by a factor of 4, and
moved 1 unit down. The resulting equation of this
conic must be;
This problem is quite interesting because we have
to apply the things that we learned in
transformation...
Expanding horizontally by a factor of 4 meaning
1
that we have to multiply the x by and
4
moving it by 1 meaning that we have to add 1.
3

4. Continuation...
x
2( )2 + 3 (y+1)2 ­ 12 = 0
4
x
( )2 6y + 3y + 6 = 12
8
x2 + 24y2 + 48y + 24 ­ 98 = 0
x2 + 24y2 + 48y ­ 72 = 0
4

5. Well, i really don't have to talk about the
questions 3 and 4 because it was pretty
easy. However, the long answer question
was very hard and i didn't even know that
we can do that such a thing...let's see. uhm...
The question is, Find the equation of the ellipse
with the end points of one axis at (3,0) and (­
22
3,0) and containing the point (2, ).
3
5

6. We do know that the formula we are going to
use is the vertical formula.
y2
2
x + 2 = 1
b2
x a
­3 3
y
Whapak! Yes, you
are right... the x, y,
and even a are
given. So just have
to solve for b...
6

7. Afternoon Class...
Well, we started talking about some interesting
topics...and then he had given some review
questions.
7

8. b = ? c = 6 a = 2
a2 + b2 = c2
b2 = c2 ­ a2
b2 = (6)2 ­ (2)2
b2 = 36 ­ 4
2
b = 32
y 2 = 1
2
x­
32
4
8

9. 2
2
x ­ y = 1
32
4
The answer was very straight forward.
Using the Pythagorean theorem, we can
easily identify the solution to our problem.
9

10. We are looking for the
equation of the
parabola.
vertex is at (­1,3)
eq'n of the directrix is
at x= 2
10

11. Continuation...
Given the vertex and it's directrix, we
can easily recognize that the parabola is
horizontal and facing to the left which
means that our p is negative.
p = ­3
vertex is at (­1,3)
(y ­ k)2 = 4p (x ­ h)
eq'n of the directrix is
at x= 2
(y ­ 3)2 = 4(­3) (x + 1)
(y ­ 3)2 = ­12 (x + 1)
11

12. 12

13. These are the points (red) that we can be
our center. Mr. K had said that all of them
are right but the easiest one is the first
one.
13

14. (x ­ h)2 = 4p (y­k)
(10 ­ 0)2 = 4p (0­40)
100 = ­160 p
­5 = p
8
­5
4p = ( )
8
= ­5
V (0, 40) 2
P (10, 0)
x2 = ­5 (30 ­ 40)
2
14

15. Whew..Finally I am done..
I can do my BOB now.
Goodluck guys! Happy
Studying..and Happy
Holidays!!!
15

16. The scribe for Friday's
class is Remyshire...
because tomorrow is our
unit test. I think that's it.
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