Parametric Equations/Parabola help (1 Viewer)

[Riviet]

Please show as much working as you can, thx.

1. A line drawn from the vertex V to the point P(x₁,y₁) on the parabola x²=4ay intersects the directrix at Z. Prove ZS (S being the focus) is parallel to the tangent at P.

2. The points P(2ap,ap²) and Q(-2a/p,a/p²) lie on the parabola x²=4ay.
a) Show that the gradient of PQ is (p²-1)/2p
b) Show that the chord PQ passes through the focus of the parabola.
c) If the midpoint of the chord PQ lies on the line 2x=3a, find the length of the chord PQ.
I got parts a) and b), so just assume the results of these two, for c) i had a go but didn't get anywhere.

3. HARD- How many normals pass through (0,Ka), a point on the axis of the parabola x²=4ay for K>2? For K=3, find where the normal meets the parabola again.

 

[Yip]

1. Since P lies on the parabola, y1=(x1)^2/4, so P(x1,(x1)^2/4a)
Line VP: y=(y1/x1)x
=(x1/4a)x
since Z lies on the directrix VP cuts Z at y=-a
substituting y=-a to find x coord of Z,
-a=(x1/4a)x
x=-4a^2/x1
so:
Z (-4a^2/x1, -a)
S(0,a)
P(x1,(x1)^2/4a)

m(SZ)=2a/4a^2/x1
=x1/2a

y=x^2/4a
y'=x/2a

m(tangent at P)=x1/2a
thus tangent at P||SZ

2.(c) Since M lies on 2x=3a, the coordinates of M are (3a/2,y)
Now, calculate the x-coordinate of M using midpoint formula
ie M's x-coord=[2ap-2a/p]/2
equate this with the given coordinate of M,
[2ap-2a/p]/2=3a/2
2ap^2-2a=3ap
2p^2-2=3p
2p^2-3p-2=0
(2p-1)(p-2)=0
p=0.5, 2
Now, to find the chord length of PQ,
substituting p=2 into P and Q,
P(4a,4a) Q(-a,a/4)
using distance formula,
D=root[25a^2-225/16a^2]
=root[625a^2/16]=25a/4
[note: u get the same result if u use p=0.5]

3.Let P(2ap,ap^2) be a point on the parabola
y=x^2/4a
y'=x/2a
m(normal)=-1/p
equation of normal: x+py=2ap+ap^3
substituting coordinate (0,Ka),
Kap=2ap+ap^3
K=2+p^2
p^2=K-2
when K>2, K-2>0
therefore there are 2 real solutions to equation and thus 2 normals pass through (0,Ka)
When K=3,
p^2=3-2=1
p=1,-1
substituting p=1 into the normal,
x+y=3a
y=3a-x
finding intersection with parabola,
3a-x=x^2/4a
x^2=12a^2-4ax
x^2+4ax-12a^2=0
(x+6a)(x-2a)=0
x=-6a, 2a
y=9a, a
so the normal intersects parabola at (-6a,9a) and (2a,a)
Substituting p=-1 into normal
x-y=-3a
y=x+3a
Finding intersection with parabola,
x^2-4ax-12a^2=0
(x-6a)(x+2a)=0
x=6a, -2a
y=9a, a
so normal intersects parabola at (6a,9a) and (-2a,a)

 

[Riviet]

Thanks Yip, that really cleared things up.