The Directrix of a Parabola

https://isaacphysics.org/questions/directrix?board=a0609375-6803-40ea-8474-06dbe9f18a3c&stage=a_level

PART C - no idea where to even start - I initially thought the centre of the circle would stay constant but then realised I had the complete wrong idea.

Reply 1

mosaurlodon

OP

10

My working for Part A and B:

Any help would be greatly appreciated.

Reply 2

Eimmanuel

Study Forum Helper

15

Original post by mosaurlodon

https://isaacphysics.org/questions/directrix?board=a0609375-6803-40ea-8474-06dbe9f18a3c&stage=a_level

PART C - no idea where to even start - I initially thought the centre of the circle would stay constant but then realised I had the complete wrong idea.

Some guides that would help.
Consider an arbitrary point say P(x,y) on the projectile path.
We know how to describe the x-coordinate and y-coordinate in terms of initial speed u, angle θ, g and t.
Use the trigo identity sin²θ + cos²θ = 1 to "transform x and y into circle equation" and we can find the centre of circle from the circle equation.

Reply 3

mosaurlodon

OP

10

I am not sure what you mean - in fact I am not entirely sure whether I understand what the q is asking.
I picked the max height to be point P, and labelled desmos to what I thought was happening.

My initial thought was to use discriminant as both graphs meet at one point but not sure if this is correct approach and how sin²θ + cos²θ = 1 relates to this.

Reply 4

mqb2766

21

Original post by mosaurlodon

I am not sure what you mean - in fact I am not entirely sure whether I understand what the q is asking.
I picked the max height to be point P, and labelled desmos to what I thought was happening.

My initial thought was to use discriminant as both graphs meet at one point but not sure if this is correct approach and how sin²θ + cos²θ = 1 relates to this.

You want to show that at any time, the locus of (x,y) values for varying theta lie on a circle. So as above use the pythagoras identity c^2+s^2=1 and start from the original equations of motion
x = vtc
y = vts - gt^2/2
and rearrange them for s and c, square and sub and ...

tbh, you should almost be able to spot the radius and x and y centres without explicitly doing the rearrangement (x~rc, y~rs). As you should expect theyre functions of t and the radius will expand as t increases (it must start at zero) and the x centre is constant but the y centre decreases as time increases. You use the pythagorean identity to eliminate the parametric variable theta (as usual).

In desmos its something like
https://www.desmos.com/calculator/4hxhy96ctx

(edited 1 year ago)

Reply 5

Eimmanuel

Study Forum Helper

15

Original post by mosaurlodon

I am not sure what you mean - in fact I am not entirely sure whether I understand what the q is asking.
I picked the max height to be point P, and labelled desmos to what I thought was happening.

My initial thought was to use discriminant as both graphs meet at one point but not sure if this is correct approach and how sin²θ + cos²θ = 1 relates to this.

If you want to know what the question is asking, see the animation in the link.
https://ibb.co/Lxj4Kkp

(edited 1 year ago)

Reply 6

Eimmanuel

Study Forum Helper

15

Original post by mosaurlodon

I am not sure what you mean - in fact I am not entirely sure whether I understand what the q is asking. [\quote]

The guides (by right should be self-explanatory) because you have done it in part A.
Consider an arbitrary point say P(x,y) on the projectile path.
We know how to describe the x-coordinate and y-coordinate in terms of initial speed u, angle θ, g and t. (see below)
The Directrix of a Parabola01_crop.jpg

I say describe the x-coordinate in terms of initial speed u, angle θ, g and t means to express the x-coordinate in terms of initial speed u, angle θ, g and t.
Have you not done in Part A?
Use the trigo identity sin²θ + cos²θ = 1 to transform x and y into circle equation …(how)
Both x and y have a trigo ratio respectively, you can make the trigo ratio the subject and then square the trigo ratio to substitute them into the sin²θ + cos²θ = 1 and transform it into a circle equation:
(x-a)^2 + (y-b)^2 = r^2.

Advice:
Learn to read the sentences slowly and repeat them a few times slowly.
Very often, students like to rush through the sentences and refuse to read them again when they find the sentences confusing.

Reply 7

mosaurlodon

OP

10

Original post by Eimmanuel

If you want to know what the question is asking, see the animation in the link.
https://ibb.co/Lxj4Kkp

Ohhhhhhh that makes sense to me now.
Actually what you're saying makes a lot more sense with that diagram - I in fact did not understand what the question was saying and had the complete wrong idea of the circle.
Ive managed to do both remaining parts now - thanks guys 😊

The Directrix of a Parabola

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