Tuesday, May 19, 2009

CoNiCs!!!!! go Mitsubishi Honda Civic!!!!!

Hey everyone, Jessi here. Welcome officially to the last month of school. Well now to get down to business.

Very first thing in class Mr. K introduced his split personality, the Samurai Mitsubishi Honda Civic. Well this very interesting individual was there to teach us all about the mystical ways of the CONIC ARTS.

1) if you take a line and spin it you create a double napped cone.

2) if you take the cone created and cut it across the top, parallel to the base, the cross section is a circle.

3) if you take the cone again and cut it across at an angle it creates an ellipse.

4) once again take the cone and cut it at an angle, cutting down to the base. The cross section of the shape created is a parabola.

5) lastly cut the double napped cone and cut it perpendicular to the base. This creates a hyperbola

Once we finished that, Mr. K had us put a dot on anywhere on in the bottom quarter of a piece of paper. We placed dashes on the edge of the paper which we then folded over to touch the dot. By doing about ten of these dashes, we found that the crease created by the folds made a parabola. We then placed a dot anywhere on the curve of the parabola. It was found that this dot was the same distance from the first dot we put on the paper and the edge of the paper. This means that a parabola is the locus of points that are of an equal distance from a fixed point and a line.

When we discussed it we found the almost all of the points on a parabola have a corresponding point which is the same distance from the focus as it. There is only one that does not. This is because it is on the vertex of the parabola meaning that it is exactly half way between the focus and the directrix. The distance between the vertex and the directix or the vertex and the focus is the same and is notated as p. This makes the equation of the directrix k-p if k is the x value of the vertex. This would also make the focus's x value equal to k+p.


since the line between P and F is equally to the line between P and D then
eq=\sqrt{(x-h)^2 +(y-(k+p))^2 } = \sqrt{(x-x)^2+(y-(k-p)^2 }

eq==y^2 - 2ky- 2py-2ky+k^2 +p^2

eq=(y-k+p)^2 = (y-k+p)(y-k+p)
eq== y^2-yk+yp-yk+k^2-kp+yp-kp+p^2
eq== y^2-2yk+2yp-2yk+k^2+p^2

eq=(x-h)^2+y^2-2ky-2py+k^2+2kp+p^2= y^2-2yk+2yp-2yk+k^2+p^2
eq=(x-h)^2= 4p (y-k)

This means that the equation for a parabola is
eq=(x-h)^2= 4p (y-k)
Because of this the equation
is equal to the equation

eq= (x-h)^2= 4p (y-k)

this means that 4p is equal to 1/a

Well that's really all that we covered...yesterdayI guess lol. Sorry for the late post everyone. See you guys in class soon.

Oh and the scribe for tommorow is...... trinhn92!!!
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