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Parametrics Q (1 Viewer)

shaon0

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P(2ap,ap^2) and Q(2aq,aq^2) are two points on the parabola x^2=4ay. The tangents TP and TQ drawn from an external point T cross the x-axis at A and B respectively.
i) Find the co-ordinates of A and B.
I know that A and B are A(x1,0) and B(x2,0) but i don't know what to do next.
 

shaon0

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Some help needed...
ii) Find the equation of the line passing through the focus that bisects the line AB.
iii) Hence, show that this line is perpendicular to the chord of contact T.
 

lyounamu

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shaon0 said:
Some help needed...
ii) Find the equation of the line passing through the focus that bisects the line AB.
iii) Hence, show that this line is perpendicular to the chord of contact T.
question?
 

shaon0

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shaon0 said:
P(2ap,ap^2) and Q(2aq,aq^2) are two points on the parabola x^2=4ay. The tangents TP and TQ drawn from an external point T cross the x-axis at A and B respectively.
i) Find the co-ordinates of A and B.
I know that A and B are A(x1,0) and B(x2,0) but i don't know what to do next.
the one above
 

lolokay

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focus is at (0,a)
AB is (ap, 0) , (aq, 0)
it's midpoint is (a(p+q)/2 , 0)

gradient of line through focus and midAB is -a/[a(p+q)/2] = -2/(p+q)
y-a = -2x/(p+q)
2x + y(p+q) - a(p+q) = 0

chord of contact is the line joining P(2ap,ap^2) and Q(2aq,aq^2).
it's gradient is a(p+q)(p-q)/2a(p-q) = (p+q)/2
(p+q)/2 * -2/(p+q) = -1

therefore they are perpendicular
 
Last edited:

-tal-

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lolokay said:
focus is at (0,a)
AB is (ap, 0) , (aq, 0)
it's midpoint is (a(p+q)/2 , 0)

gradient of line through focus and midAB is -a/[a(p+q)/2] = -2/(p+q)
y-a = -2x/(p+q)
2x + y(p+q) - a(p+q) = 0

what's a chord of contact?
A chord of contact is the line in between the parabola that touches points Q & P.
 

Continuum

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P: y=px-ap2
Q: y=qx-aq2
px-ap2=qx-aq2
Rearrange to get: x=a(p+q)
Sub in x into one of the equations to get: y=-apq
T: a(p+q),-apq
Chord of contact: xx0=2a(y+y0)
ax(p+q)=2a(y-apq)
Rearrange to form y=(p+q)/2 . x + apq
Gradient of chord of contact: (p+q)/2
(p+q)/2 . -2/(p+q) = -1
Therefore perpendicular
 

shaon0

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1. If the chord of contact to the parabola x^2=4ay from an external point P(x0,y0) passes through the point A(0,2a), show that the locus of the midpoint of PA is the x-axis.
.....I have the midpoint of PA, what should i do next?
2.The tangent at the point P(2ap,ap^2) on the parabola x^2=4ay cuts the y-axis at T. The line via the focus S parallel to this tangent cuts the directrix at V. M is the midpoint of TV. Find the locus of M as P moves on the parabola.
 

NitroBoon

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To determine the locus of a parametric point, all you have to do is to eliminate the parameter(s) from the coordinate
 

lolokay

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in 1. you should get pq = -2, and y0 = apq = -2a, so the midpoint of 2a and -2a is the x axis
 

NitroBoon

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shaon0 said:
2.The tangent at the point P(2ap,ap^2) on the parabola x^2=4ay cuts the y-axis at T. The line via the focus S parallel to this tangent cuts the directrix at V. M is the midpoint of TV. Find the locus of M as P moves on the parabola.
Solution:


<o></o>Feel free to post corrections, I am not too sure about the answer lol, seems abit random.
 

shaon0

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NitroBoon said:
Solution:


<o></o>Feel free to post corrections, I am not too sure about the answer lol, seems abit random.
thanks for the solytion. i got most of the question write in the beginning but i made a stupid mistake on expanding and getting the Locus of M.
Thanks :)
 

Aerath

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How do you do a solution on the computer like that? :p
 

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