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Revision of locus and parabola + Sequence and series (2 Viewers)

fullonoob

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A(3,2) and B(-3,4). AP is perpendicular to BP. Find the equation of P(x,y).
using m1m2 = -1
y-4/ x +3 x y-2/ x-3 = -1
(y-4)(y-2) = - (x-3)(x+3)
rearranging, you get x^2 - y^2 - 6y - 1 =0
which is a circle
 

fullonoob

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my turn
find a point on the parabola x^2 = 4ay which has its normal passing through its focus. Prove that this is the point without simply stating it.
 

hscishard

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wrong there is one.
Can anyone prove this?
Instead of simply stating stuff like carrot :lol:
Sibilance! Too bad its a maths topic.
Ok so what?
Differentiate that = x/2a
So y= -2a/x(x) + a
...I dont get it how do you prove it? Do you just simple substitute 0,a?
 

Carrotsticks

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if you must be that picky.....

it is the normal at the vertex.
 

fullonoob

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if you must be that picky.....

it is the normal at the vertex.
rofl didn't i say prove it, you get 0 for stating it. It is obvious that it is at (0,0) as i said "one point". Anyone else wna try being a genius like carrot, maybe apply e = mc^2 along the way? :hat:
 

Carrotsticks

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why do you go on the 2U forums posting 3U stuff anyway? Does it make you feel smart that you're so 'good' compared to 2U people? Typing maths on the computer is a very tedious task, so I'd prefer just to state the method.

If I see you IRL, I will no longer state, and will do the work =)
 

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