Mimi_x3
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
where is point S?
Mimi_x3
  • Mimi_x3
Um, idk, I think S is the focal length
anonymous
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Is this what it asks?
Mimi_x3
  • Mimi_x3
idk, i don't understand the question, i just don't understand parametrics xD
Mimi_x3
  • Mimi_x3
Can you help ?
anonymous
  • anonymous
I don't know what the question is asking for.
anonymous
  • anonymous
You must provide more details.
Mimi_x3
  • Mimi_x3
Oh yeah I forgot sorry a) Prove (i) OS=OA (O is vertex, S is the focus)
anonymous
  • 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
  • Mimi_x3
Don't you have to solve it or something to prove that OS=OA like finding the equation of tangent
anonymous
  • anonymous
for the question a) OS=OA, I don't think so.
anonymous
  • anonymous
i think you will use it when you are going to solve other questions
Mimi_x3
  • Mimi_x3
Okay, I will draw the diagram that I drew can you check if its correct ?
anonymous
  • anonymous
no problem.
Mimi_x3
  • Mimi_x3
|dw:1318745104246:dw|
Mimi_x3
  • Mimi_x3
Is it right ?
anonymous
  • 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
  • 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
  • anonymous
for now, you don't need to worry about the other P, just to solve the question will be OK.
Mimi_x3
  • Mimi_x3
Okay, how can I solve it ?
anonymous
  • 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
  • anonymous
PS=PN, P is the point on the parabola----the definition
Mimi_x3
  • Mimi_x3
Okay thank you. But one more how can I prove ON=OT, do I use the same definition ?
anonymous
  • 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
  • Mimi_x3
wow~ thank you. I drew it totally wrong *sigh* I just don't understand parametrics.
anonymous
  • 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
  • Mimi_x3
Yeah, I should practice on it. How would I know if its vertical or horizontal ?
anonymous
  • anonymous
X^2=something*Y is vertical, Y^2=something*X is horizontal.
Mimi_x3
  • Mimi_x3
Oh I get it, thank you very much ^_^
anonymous
  • anonymous
You are welcome. I am a PHD student, so heyhey.
Mimi_x3
  • Mimi_x3
Cool~ PHD student. Wonder why you're smart.
anonymous
  • 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
  • Mimi_x3
Lols, I'm afraid it esp parametrics, its like so~ HARD.
Mimi_x3
  • Mimi_x3
of it*
anonymous
  • 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
  • Mimi_x3
Oh cool~
anonymous
  • 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
  • Mimi_x3
Okay, I will work it out.

Looking for something else?

Not the answer you are looking for? Search for more explanations.