Parametrics Question (1 Viewer)

[rama_v]

Parabola Question I can't get.

H (2ah, -ah^2) and Q (2aq, -aq^2) are two variable points on the parabola x^2=-4ay

Let the point of intersection of HQ with the y axis be the fixed point T (0, -3a) where a>0. The normals to the parabola at the points H and Q meet at the point K. Assume that h>0 and q<0.

i) Show that qh=-3 ( I got this part)
ii) Show that the coordinates of the point K are:
(3ah - 9a/h , a - ah^2 - 9a/h^2)

 

[KFunk]

ii) First, find the equation of the normal:

dy/dh = -2ah dx/dh=2a ---> dy/dx = -h therefore gradient of normal = 1/h

y +ah² = 1/h(x-2ah)
x = hy + 2ah + ah³
and
x = qy + 2aq + aq³

hy + 2ah + ah³ = qy + 2aq + aq³
y(h-q) = -2a(h-q) -a(h³ - q³)
y = -2a -a(h² +hq + q²) where q=-3/h (as you have proven)
y= -2a -a(h² -3 + 9/h²)
y= a - ah² - 9a/h² as required

EDIT: I was going to leave it there and go sleep but I figured I may as well answer it in full.

x = hy + 2ah + ah³
x = h(a - ah² - 9a/h²) + 2ah + ah³
x = ah - ah³ -9a/h + 2ah + ah³
x= 3ah - 9a/h

since x= 3ah - 9/h and y= a - ah² - 9/h² the point of intersection is at:
K(3ah - 9a/h , a - ah² - 9a/h²)

 

[rama_v]

Thanks