Antoinne:
Corrected. thanks

agentofchange1:
#just. drinking. Garri and observing


happy solving guys...

@ Laplacian ( m ), Karmanaut greetings in CAPITAL letters

agentofchange1:
#just. drinking. Garri and observing


happy solving guys...

@ Laplacian ( m ), Karmanaut greetings in CAPITAL letters

Laplacian:
i feel ur efforts bro, kudos!!...enjoy urself!!!

Karmanaut:
My oga, I hail.

jackpot:
please, help me out.


2. Consider the hyperbola 4x^2-4x-y^2+6y+28=0. Obtain
i. Eccentricity
ii. Foci
iii. Vertex
iv. Centre
v. The Length of it's semi-latus rectum
vi. It's asymptotes.[/b]

tags: agentofchange1, dejt4u, Laplacian, Antoinne, Karmanaut, doubleDx, etc

jackpot:
.

Foci: (1/2, 3-√5); (1/2, 3+3√5)
Vertices: (1/2, -3); (1/2, 9)
Center: (1/2, 3)
Semi major axis: 6
Semi minor axis: 3
Focal parameter: 3/√5
Eccentricity: √5/2
Asymptote: -6x/[√(1/4 (-3×3√5 + 3(1+√5))^2 -36)]
+ 3/[√(1/4(-3+3√5 +3(1+√5))^2 -36)] +3
Which is simply 4x-2, forgive my obfuscation.

Focus: √(a^2 + b^2) ; (-sqrt(a^2+b^2)
focal parameter: b^2/sqrt(a^2+b^2)
eccentricity: sqrt(b^2/a^2+1)
asymptotes: y = (b x)/a; y = -(b x)/a 

Karmanaut:


PS: Seun, we need LaTeX on Nairaland.

Karmanaut:

 Asymptote: -6x/[√(1/4 (-3×3√5 + 3(1+√5))^2 -36)]
 + 3/[√(1/4(-3+3√5 +3(1+√5))^2 -36)] +3

The other asymptote is the conjugate of this.

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ayodijex:


...i thought assymptote is simply +/-(b/a)??.

Antoinne:

You wish he could do that!

Antoinne:
So, I have a problem that occured to me while solving the parabola problem earlier. Don't know if you guys are interested

Given a parabola, a circle with radius the length of parabola focus is placed at a point N on the parabola, n distance away from the center of the parabola. If this circle falls along the parabola and rises to the other end of the parabola, equally n distance away to the right, how many revolutions will it make? You can express answer in terms of all constants.

I'm wondering this may have some application with planetary bodies

tags: agentofchange1, dejt4u, Laplacian, Antoinne, Karmanaut, doubleDx, etc

Laplacian:

all things being equal, the equation of the parabola is;
4ay=x², now the length of any curve is given by;
S=§√[1+(y')²]dx, in our case, y'=x/2a, integrate from x=-n to x=n,
S=§√[1+(x/2a)²]dx, substitute x=(2a)tan@, the integration should be easy for you!! The length of the circle is
L=2#a, so the number of revolutions N=S/L.
Am lukin for a concise way of solving Jackpot's second problem without rotating the axis.

Antoinne:


Good work. You on track. I was hoping you'd give the full solution, though. So, we can see the relationships clearly and see if it offers any insights (hyperbolic or parabolic trigonometric functions).
Good try still.

Laplacian:

ok. I was in a car then.
Let k=√[1+(2a/n)²] then,
S=2aLog[(k+1)/(k-1)]

N=S/L=(1/#)Log[(k+1)/(k-1)].
An important deduction is that no circle can ever make whole number of revolutions

Antoinne:
you sure about this?

Laplacian:

100%

chezholy:
1. I. 9.3cm
II.24cm
III. 201.81cm²

chezholy:
1. I. 9.3cm II.24cm III. 201.81cm²

Geofavor:

2) PQRS is a parallelogram. The distance between PS and QR is 14cm and PS = 25cm. If PQ = 17.5cm find the distance between PQ and SR.

dejt4u:
You are correct

Geofavor:

I too couldn't fault that working. But this textbook has the answers to be different.
I) 12cm
Ii) 30cm
Iii) 336cm^2

dejt4u:

4cm

chezholy:
1. I. 9.3cm
II.24cm
III. 201.81cm²

dejt4u:
the textbook can be very wrong when it comes to maths.. Cheers