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Mimi_x3
 5 years ago
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
Mimi_x3
 5 years ago
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

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Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.4Um, idk, I think S is the focal length

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this what it asks?

Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.4idk, i don't understand the question, i just don't understand parametrics xD

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know what the question is asking for.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You must provide more details.

Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.4Oh yeah I forgot sorry a) Prove (i) OS=OA (O is vertex, S is the focus)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then 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
 5 years ago
Best ResponseYou've already chosen the best response.4Don't you have to solve it or something to prove that OS=OA like finding the equation of tangent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the question a) OS=OA, I don't think so.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think you will use it when you are going to solve other questions

Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.4Okay, I will draw the diagram that I drew can you check if its correct ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uhh, 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
 5 years ago
Best ResponseYou've already chosen the best response.4hmm..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
 5 years ago
Best ResponseYou've already chosen the best response.0for now, you don't need to worry about the other P, just to solve the question will be OK.

Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.4Okay, how can I solve it ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which 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
 5 years ago
Best ResponseYou've already chosen the best response.0PS=PN, P is the point on the parabolathe definition

Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.4Okay thank you. But one more how can I prove ON=OT, do I use the same definition ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, 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
 5 years ago
Best ResponseYou've already chosen the best response.4wow~ thank you. I drew it totally wrong *sigh* I just don't understand parametrics.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0parabola 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
 5 years ago
Best ResponseYou've already chosen the best response.4Yeah, I should practice on it. How would I know if its vertical or horizontal ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0X^2=something*Y is vertical, Y^2=something*X is horizontal.

Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.4Oh I get it, thank you very much ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are welcome. I am a PHD student, so heyhey.

Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.4Cool~ PHD student. Wonder why you're smart.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank 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
 5 years ago
Best ResponseYou've already chosen the best response.4Lols, I'm afraid it esp parametrics, its like so~ HARD.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i see. Parametrics is harder but you can consider of them as changeable constants. They are constants but they are changing in different situation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0back 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=(12a)*P^2.

Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.4Okay, I will work it out.
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