- Mimi_x3

P is the point (2ap,ap^2) on the parabola x^2=4ay. The tangent at P meets the axis of the parabola at T and PN is drawn perpendicular to the axis, meeting it at N. The directrix meets the axis at A.
(a) Prove OS=OA

- katieb

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- anonymous

where is point S?

- Mimi_x3

Um, idk, I think S is the focal length

- anonymous

the definition of the parabola is that the distance between the point P with the focus S is equal to the distance from the point P to the linear directrix.
OS=OA is the special situation when P is point O.

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## More answers

- anonymous

Is this what it asks?

- Mimi_x3

idk, i don't understand the question, i just don't understand parametrics xD

- Mimi_x3

Can you help ?

- anonymous

I don't know what the question is asking for.

- anonymous

You must provide more details.

- Mimi_x3

Oh yeah I forgot sorry
a) Prove (i) OS=OA (O is vertex, S is the focus)

- anonymous

then i have typed the answer:
the definition of the parabola is that the distance between the point P with the focus S is equal to the distance from the point P to the linear directrix.
OS=OA is the special situation when P is point O since point O is on the parabola.

- Mimi_x3

Don't you have to solve it or something to prove that OS=OA like finding the equation of tangent

- anonymous

for the question a) OS=OA, I don't think so.

- anonymous

i think you will use it when you are going to solve other questions

- Mimi_x3

Okay, I will draw the diagram that I drew can you check if its correct ?

- anonymous

no problem.

- Mimi_x3

|dw:1318745104246:dw|

- Mimi_x3

Is it right ?

- anonymous

uhh, actually, the directrix is Y=-a, not PN, so point A should be (0,-a) not (2ap, 0)
and I don't know why there is another point P on the left side.

- Mimi_x3

hmm..how does it look like then ?
In my textbook, in one of the examples there's another P as well so I assume that there's another P in this question idk

- anonymous

for now, you don't need to worry about the other P, just to solve the question will be OK.

- Mimi_x3

Okay, how can I solve it ?

- anonymous

which question? question a is solved through definition of parabola by using the special situation when the point P on the parabola moves to the vertex which is point O.

- anonymous

PS=PN, P is the point on the parabola----the definition

- Mimi_x3

Okay thank you. But one more how can I prove ON=OT, do I use the same definition ?

- anonymous

OK, there is one more mistake which is the axis is X=0 not Y=0 since this parabola is vertical not horizontal.
let me draw another diagram just to let you know what it should be.
|dw:1318746374530:dw|

- Mimi_x3

wow~ thank you. I drew it totally wrong *sigh* I just don't understand parametrics.

- anonymous

parabola can be vertical or horizontal which depends on the format of the function. The best way to differentiate which is which for me is to learn how to draw it correctly first.

- Mimi_x3

Yeah, I should practice on it. How would I know if its vertical or horizontal ?

- anonymous

X^2=something*Y is vertical, Y^2=something*X is horizontal.

- Mimi_x3

Oh I get it, thank you very much ^_^

- anonymous

You are welcome. I am a PHD student, so heyhey.

- Mimi_x3

Cool~ PHD student. Wonder why you're smart.

- anonymous

Thank you. I am just happy to help people especially in Mathematics because I love it. Don't be afraid of it. It is really cool when you see the essence inside of it.

- Mimi_x3

Lols, I'm afraid it esp parametrics, its like so~ HARD.

- Mimi_x3

of it*

- anonymous

i see. Parametrics is harder but you can consider of them as changeable constants. They are constants but they are changing in different situation.

- Mimi_x3

Oh cool~

- anonymous

back to the question ON=OT, you can not use definition since they are not distances in definition. You need to solve the tangent equation and find out point T. I found it. but they are not equivalent. ON=P^2, OT=(1-2a)*P^2.

- Mimi_x3

Okay, I will work it out.

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