### azn_chilly's scribe post (problems + learning)

This morning we started off with three questions off the bored and i was late for that class and was who willing to the scribe today hahaha!!!!!!. so we stared off by this .....

graph this conic section and label all information points

4x(square)+9x(square)-16x+18y-11=0

4x(square)-16x + 9x(square)+18y =11

(4x(square)-16x+4-4) + (9x(square)+18y+1-1)=11+16+9

4(x-2)(square)/36 + 9(y+1)(square)/36 = 36/36

(x-2)(square)/9 + (y+1)(square)/4 =1

(x-2)(square)/3 + (y+1)(square)/2 =1

A(square) - B(square)=C(square)

3(square) - 2(square)=C(square)

9-4=C(square)

5=C(square)

(square root)5=C

find the equation of an ellipse with center at (2,-5), major axis at length 6 units parallel to the x-axis and major axis 2 units long

(x-2)(square)/9 - (y+5)(square)/1=1

a tunnel has a semi-elliptical cross-section it is 20m wide and 5m at its highest point, find the equation of the ellipse and use it to find the height of the tunnel 2m away from one end

(x-h)(square)/a(square) + (y-k)(square)/b(square)=1

x(square)/100 + y(square)/25=1

8(square)/100 + y(square)/25=1

y(square)/25=100/100-64/100

y(square)/25=36/100

(25)y(square)/25=36/100(25)

y(square)=9

y(square)-9=0

(y+3)(y-3)=0

y=-3 y=3

y=-3 is rejected cause it will be in the negative side

in the afternoon we talked about of the construction of the hyperbola and its properties...

(x-h)(square)/a(square) - (y-k)(square)/b(square)=1 is the hyperbola

(x-h)(square)/a(square) + (y-k)(square)/b(square)=1 is the ellipse

hyperbola: a locus of points that moves in such a way so that the absolute value of the distance(focal radii) from two fixed points(foci) is constant

|PF1-pf2|= 2a[a constant]

the anatomy of a hyperbola

0 is the centre

A1 + A2 are the verticies

A1A2 is the transverse axis its length is 2a

B1 +B2 are the end point of the conjugate axis its length is 2b

F1 + F2 are the foci they are for the centre

PF1 and PF2 are the focal radii

|PF1 - PF2|= 2a

the picture of the hyperbola is horizontal and centred at the the origin, the equation of the of the asymptotes are: y= b/ax and y=-b/ax

the transvers and conjugate axis are lines of symmetry for the hyperbola

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