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another conics Q (1 Viewer)

sincred91

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The normal at a point P(at^2,2at) on the parabola y^2=4ax intersects the x axis at Q. Find in terms of t, the coordinates of Q. Find also the equation of the locus of the mid-point of PQ. Show taht the locus is a parabola for all t doesnt equal 0

ah so hard.
 

hermand

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The normal at a point P(at^2,2at) on the parabola y^2=4ax intersects the x axis at Q. Find in terms of t, the coordinates of Q. Find also the equation of the locus of the mid-point of PQ. Show taht the locus is a parabola for all t doesnt equal 0

ah so hard.
find the equation of the normal by first finding the gradient of the tangent, then the gradient of the normal is the negative reciprocal. then sub gradient and the point P into y-y1=m(x-x1)

then put y=0 in and find the x value, therefore Q will be ('x-value',0).

for the first question, for the second, i'll leave that to someone else. i'm too sick and tired to think.
 
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have u got the answer? i'm not sure if this is right.. i hope it is :)

d/dx (y^2) = d/dx (4ax)
2y.(dy/dx) = 4a
dy/dx = 4a/2y
i.e. dy/dx = 2a/y

At P (at^2, 2at) -> dy/dx = 2a/2at = 1/t
i.e. gradient of normal will be -t

y-y1 = m(x-x1)
y - 2at = -t (x - at^2)
y - 2at = -tx + at^3

For Q, let y = 0

i.e. tx = at^3 +2at
x = at^2 + 2a

i.e. Q is (at^2+2a, 0)

For midpt of PQ: Let it be (x,y)
x = (at^2 + at^2 + 2a ) / 2 = at^2 + a ------1
y = (2at + 0) / 2 = at ---- --2

from 2 :
t = y/a
sub in 1

i.e. x = a (y/a)^2 +a
= y^2/a + a
i.e. a(x-a) = y^2 is the locus

Parabola vertex on (a,0) , focal length a/4

hmm i don't know why t can't equal 0... maybe it's because if t is 0, then p is zero and then the normal at P can't really intersect the x-axis at a distinct pt Q

i think my teacher said conics was just a name for the family of curves :hyperbola, ellipse, circle and parabola, but tbey call it parametrics in the 3u syllabus
 
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