Focus in parametrics (2 Viewers)

[Help Galore]

In trying to do some locus problems, some of the questions which involve the focus (0, a) tend to switch the 'a' with a 1. I can't see how this works. I highly doubt my answer is wrong, but I'm pretty sure the answer in the back is also correct. It's probably a stupid question, but I would like to know.
Thanks in advance.

 

[ohexploitable]

I don't understand what you're trying to say, post the question pl0x.

 

[Help Galore]

P(2p,p^2) is a variable point on the parabola x^2 = 4y, whose focus S and vertex O. M, N are the midpoints of SP, OM respectively.
(i) Show the locus of M is the parabola y = 1/2(x^2 + 1), and find the locus of N.

That isn't the full question. It's from Coroneos 3U btw. I don't know why I find Coroneos more difficult then the Fitz in parametrics >_>.

 

[random-1006]

P(2p,p^2) is a variable point on the parabola x^2 = 4y, whose focus S and vertex O. M, N are the midpoints of SP, OM respectively.
(i) Show the locus of M is the parabola y = 1/2(x^2 + 1), and find the locus of N.

That isn't the full question. It's from Coroneos 3U btw. I don't know why I find Coroneos more difficult then the Fitz in parametrics >_>.

thats not that hard, ill post up answer now


Focus= S= ( 0,1)
Vertex = O= (0, 0)

M is midpoint of SP, ie x= (0 +2p) / 2= p
y= (p^2 +1) /2 = (x^2 +1) / 2

ie M ( p, (p^2 +1)/ 2)

N is midpoint OM:

x= (p +0)/2, ie p=2x
y= [(p^2 +1) / 2 + 0 ] / 2
= [p^2 + 1 ]/ 4
= [4x^2 +1] /4 ----> locus of N

 

[Help Galore]

It's not necessarily the difficulty of the question. I'm just confused as to why the a becomes a 1.

 

[MetroMattums]

It's not necessarily the difficulty of the question. I'm just confused as to why the a becomes a 1.

1 is just a constant substituted in place of the focal length (a)?

 

[hscishard]

Are you saying A1?

Or do you mean a is 1?

a is the length between the focus and the vertex = length between vertex and directrix.

In your example, x^2 = 4y
Comparing to x^2 = 4ay, a = 1.

 

[random-1006]

It's not necessarily the difficulty of the question. I'm just confused as to why the a becomes a 1.

equate x^2 = 4ay with x^2 = 4y

 

[Help Galore]

*sigh*, I knew it was something simple -.-

Thanks anyways, sorry to bother you with a stupid question

 

[ohexploitable]

Standard form is x²=4ay, the given parabola is x²=4y, so,

4ay=4a
4a=4
a=1

 

[Help Galore]

Can I post up another question?

 

[ohexploitable]

Sure...

 

[hscishard]

The a =1 question wasn't stupid.

Can I post up another question?

But this one is.

 

[Help Galore]

The tangent at P(2ap,ap^2) to the parabola x^2 = 4ay meets the x-axis at A. Find A. S is the focus of this parabola. Prove that SA is perpendicular to AP. Prove that the equation of the locus of the centre C of a circle through P,S, and A is the parabola x^2 = 2ay - a^2

 

[ohexploitable]

Solve equation of tangent and x-axis simultaneously,

y=px-ap²
y=0

0=px-ap²
x=ap
y=0

.: A(ap,0)

 

[ohexploitable]

Solve equation of tangent and x-axis simultaneously,

y=px-ap²
y=0

0=px-ap²
x=ap
y=0

.: A(ap,0)

 

[MetroMattums]

For the second part, the centre C of the circle is the midpoint M of P and S (circle geometry) - you can derive the locus from there.

 

[ohexploitable]

Gradient of SA,

m=(0-a)/(ap-0)
m=-1/p

Gradient of AP,

m=p

-1/p*p=-1

.: SA perp. to AP

 

[hscishard]

For the second part, the centre C of the circle is the midpoint M of P and S (circle geometry) - you can derive the locus from there.

I tried that but it's soo fucking long.
Unless I'm doing it wrong. We use the gen eqn of the circle right?

 

[Help Galore]

I've also gotten what ohexploitable did, and they were the right answers. A little clueless as to how to attack the circle thing though.