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