Mathematics Question Thread

[teapancakes08]

You can't treat these log ones like a geometric series. Because if you try putting everything inside one log you get this:
log(3*6*12*...)

Notice how it's all multiply. Not add. This is a product, not a sum.

Have to decompose it into
(log 3) + (log 3 + log 2) + (log 3 + 2 log 2) + ... which is an arithmetic series.

Oh! No wonder it wouldn't divide nicely...

Thanks ^^

[teapancakes08]

LAST SERIES. PROMISE.

(I'm so sorry for flooding the thread with so many of these...)

30. The sum of the first 8 terms of a geometric series 17 times the sum go its first 4 terms. Find the common ratio.

[RuiAce]

LAST SERIES. PROMISE.

(I'm so sorry for flooding the thread with so many of these...)

30. The sum of the first 8 terms of a geometric series 17 times the sum go its first 4 terms. Find the common ratio.

[katnisschung]

insert two numbers between 64 and 27 so that the sequence of four numbers is
in a geometric sequence

[RuiAce]

insert two numbers between 64 and 27 so that the sequence of four numbers is
in a geometric sequence

[f_tan]

Can anyone help me with this question?

The rate of change of V with respect to t is given by dV/dt = (2t-1)². If V=5 when t=1/2, find V when t=3.

Thanks!

[RuiAce]

Can anyone help me with this question?

The rate of change of V with respect to t is given by dV/dt = (2t-1)². If V=5 when t=1/2, find V when t=3.

Thanks!

[Thebarman]

Hey guys, couple of questions from a previous exam paper.

Q1) A series is given by:  4   +   8   +   16   + …. . The nth term of this series is given by:
A)    4 + (n  -  1)  x  4   
B)   4  x  2^n   
C)   4  x  2^n+1   
D)   2^n+1
Why is the answer D?

2) How do you describe the locus of the point, P (x,y) if it has the equation (x-3)^2 = 4(y-3). (worth 1 mark)

3) For the parabola given by the equation below, find:
x^2  +   6x  -  8y  =  -1

i)   the co-ordinates of the vertex.   
ii)   the co-ordinates of the focus.   (mainly confused about this)


Thanks!

[RuiAce]

Hey guys, couple of questions from a previous exam paper.

Q1) A series is given by:  4   +   8   +   16   + …. . The nth term of this series is given by:
A)    4 + (n  -  1)  x  4   
B)   4  x  2^n   
C)   4  x  2^n+1   
D)   2^n+1
Why is the answer D?

2) How do you describe the locus of the point, P (x,y) if it has the equation (x-3)^2 = 4(y-3). (worth 1 mark)

3) For the parabola given by the equation below, find:
x^2  +   6x  -  8y  =  -1

i)   the co-ordinates of the vertex.   
ii)   the co-ordinates of the focus.   (mainly confused about this)


Thanks!

__________________

This is just a parabola. Clearly, its vertex is at (3,3) and its focal length is a=1. So (bearing in mind its orientation) this is a parabola with focus at (3,4) and directrix y=2
__________________

[Thebarman]

Thanks!

How would I solve this? I'm getting everything but the constant right.
Find the equation of the parabola with focus (-1,3), directrix y=-1.
Focus (-4,1) and directrix y=-1.

Thanks again

[RuiAce]

Thanks!

How would I solve this? I'm getting everything but the constant right.
Find the equation of the parabola with focus (-1,3), directrix y=-1.
Focus (-4,1) and directrix y=-1.

Thanks again

Here is the method.

Do a free-hand sketch clearly showing the focus and the directrix. Then, figure out the orientation of the parabola (does it concave upwards, downwards or sideways). Then work out the focal length, and finally work out the vertex (which you can write its equation afterwards).

Note: The focal length is the distance from the vertex to the focus.
The (perpendicular) distance from the focus to the directrix is twice the focal length


If it doesn't come out, post up your working

[SSSS]

Hey. I'm having issues with this question:

Let R(1/5,-2) be a point on the parabola y^2 = 20x.
(a) Find the equation of the focal chord passing through R.
(b) Find the coordinates of the point Q where this chord cuts the directrix.
(c) Find the area of DOFQ where O is the origin and F is the focus. (d) Find the perpendicular distance from the chord to the point P^-1, -7h.
(e) Hence find the area of DPQR.

Thanks!

[RuiAce]

Hey. I'm having issues with this question:

Let R(1/5,-2) be a point on the parabola y^2 = 20x.
(a) Find the equation of the focal chord passing through R.
(b) Find the coordinates of the point Q where this chord cuts the directrix.
(c) Find the area of DOFQ where O is the origin and F is the focus. (d) Find the perpendicular distance from the chord to the point P^-1, -7h.
(e) Hence find the area of DPQR.

Thanks!

Break it down.

a) Where is the focus? It is at (0,4).
And then you should realise that this is your classic find the gradient, then sub into y-y₁=m(x-x₁) problem

b) What is the equation of the directrix? y=-4. Question: WHY is it y=-4
And then what are you doing? You're just finding a point of intersection. So whatever your answer was for part a) just sub y=-4 in.


Check if you made a typo from here on. Because you haven't told us what D is.

Also, for a question like this a freehand diagram (or computer simulation) is highly recommended, if you haven't been provided one.

[SSSS]

Here is the method.

Do a free-hand sketch clearly showing the focus and the directrix. Then, figure out the orientation of the parabola (does it concave upwards, downwards or sideways). Then work out the focal length, and finally work out the vertex (which you can write its equation afterwards).


Yeah, I made quite a few typos.
C is actually find area of traiangle OFQ where o origin and f is focus
E is find area of triangle PQR

But after your explanation, it makes sense. I fear locus because I wasn't taught it properly [missed class so whole topic was kind of taught by students]. But I'll try it again. Thanks again!

Note: The focal length is the distance from the vertex to the focus.
The (perpendicular) distance from the focus to the directrix is twice the focal length


If it doesn't come out, post up your working

[RuiAce]

If you think about it, all that falls under the locus topic is just the focus and the directrix. Everything else is coordinate geometry.

With that being said, yeah have a go. You're welcome to come back if you remain stuck (or of course a new question), but post your working if you do