The Final Blog Yes!

Morning Class... 1

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

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

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

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

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

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

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

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

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

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

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

(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

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

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

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The Final Blog Yes!

by luispc40sf07 on Dec 19, 2007

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The Final Blog Yes! — Presentation Transcript