Finding Common Ratio of Geometric Progression for 3 Points on a Parabola

[thereddevils]

Homework Statement

(p,a) , (q,b) and (r,c) are the coordinates of three points on the parabola y^2=3x. If the x-coordinate for these three points form a geometric progression whereas the corresponding y-coordinate form an arithmetic progression, find the common ratio of the geometric progression.

Homework Equations

The Attempt at a Solution

q^2=rp and 2b=c+a

Substitute those points into the parabola to get a^2=3p , b^2=3q and c^2=3r

q^2=rp

(b^4/9)=((a^2c^2)/9)

b^4=a^2c^2

b^2= +/- ac

This part confuses me. I would get a=c if i take it to be positive.

Then substituting into 2b=c+a, that would be b=a=c ?

 

[Mentallic]

You don't even have to take the positive,

from b^2=\pm ac
then b=\pm\sqrt{ac} (since we can only take the root of a positive number, we scrap the b^2=-ac result).

Substituting that into the arithmetic progression,

\pm 2\sqrt{ac}=a+c

4ac=a^2+2ac+c^2

a^2-2ac+c^2=0

(a-c)^2=0

thus a=c (and this is without assuming positive and negative cases, this is for both).

Right, so we get a=b=c. This is just telling us that it is impossible to have the criteria the question mentioned without the numbers all being equal. So what is the geometric ratio?

 

[thereddevils]

You don't even have to take the positive,

from b^2=\pm ac
then b=\pm\sqrt{ac} (since we can only take the root of a positive number, we scrap the b^2=-ac result).

Substituting that into the arithmetic progression,

\pm 2\sqrt{ac}=a+c

4ac=a^2+2ac+c^2

a^2-2ac+c^2=0

(a-c)^2=0

thus a=c (and this is without assuming positive and negative cases, this is for both).

Right, so we get a=b=c. This is just telling us that it is impossible to have the criteria the question mentioned without the numbers all being equal. So what is the geometric ratio?

Thanks Mentallic, so would it be -1? How do i show that in my working?

 

[Mentallic]

No no. We've already proven that for x coords to be in a geometric progression and the corresponding y coords to be in an arithmetic progression, where the x and y coords lie on the parabola y²=3x, then we can only have that a=b=c which means all the coordinates much be the same.

The difference in the arithmetic progression of a,b,c must be 0 (since they're all equal, b-a=0 and c-b=0)
The ratio in the geometric progression p,q,r must then be 1 since a=b=c which says that all the points are equal, so p=q=r. Thus q/p=1 and r/q=1.

 

[thereddevils]

No no. We've already proven that for x coords to be in a geometric progression and the corresponding y coords to be in an arithmetic progression, where the x and y coords lie on the parabola y²=3x, then we can only have that a=b=c which means all the coordinates much be the same.

The difference in the arithmetic progression of a,b,c must be 0 (since they're all equal, b-a=0 and c-b=0)
The ratio in the geometric progression p,q,r must then be 1 since a=b=c which says that all the points are equal, so p=q=r. Thus q/p=1 and r/q=1.

Isn't that the ratio of a GP not equal to 1 and 0 ?

 

[Mentallic]

I don't understand what you just said...

If the terms in a geometric series are equal, then the ratio is 1 since we are multiplying by 1 each time. It's the only way it can be both a geometric series and an arithmetic series at the same time.

 

[thereddevils]

I don't understand what you just said...

If the terms in a geometric series are equal, then the ratio is 1 since we are multiplying by 1 each time. It's the only way it can be both a geometric series and an arithmetic series at the same time.

oh, from the formula of the sum of a GP, Sn=(a(1-r^n))(1-r) where r not equal 1. I was referring to this which is not at all our case. Now i understand, thanks Mentallic for your help.

 

[Mentallic]

Oh, that. Well yeah that formula works for r\neq 1. For r=1 we can quickly conclude that it doesn't have to be seen as a geometric sequence anymore, but an arithmetic sequence as we've already seen. So the sum would obviously be S_n=an