dy/dx = x/2alyounamu said:P & Q are points on the parabola x^2 = 4ay with parameter value t= 1 and t=2. Show that the normals to the parabola Q and P intersect at a point R on the parabola.
Thanks in advance.
Thank you so much!3unitz said:dy/dx = x/2a
mnormal = -2a/x
y - y0 = -2a/x0 . (x - x0)
normal at P (2a, a):
y - a = -2a/2a (x - 2a)
y - a = -x + 2a
y = -x + 3a ----------(1)
normal at Q (4a, 4a):
y - 4a = -2a/4a (x - 4a)
y - 4a = -x/2 + 2a
y = - x/2 + 6a ----------(2)
equate (1) and (2):
-x + 3a = -x/2 + 6a
-x/2 = 3a
x = -6a
sub into (1) to get y:
y = 9a
.'. R = (-6a, 9a)
check if lies on parabola x^2 = 4ay:
(-6a)^2 =? 4a(9a)
36a^2 =? 36a^2
true, therefore normals intersect at R which lies on the parabola.
I already did those questions. They were worth 1 mark and 2 marks respectively. I think the question I am asking is one mark but I want to know how to do this. Anyone?u-borat said:wat does it mean by 'finished' in parts 1) and 2)?
sweet man, I will try that.u-borat said:lol my bad, i thought they were part of the question.
with the question,
as t---> infinity, e^-t-->0.
therefore, x approaches 10 from below.
then, just sub x=10.01 and solve for t.