Locus Question (1 Viewer)

[Dreamerish*~]

In the diagram above, P(2ap, ap²) is a point on the parabola x² = 4ay. The point Q lies on PS produced such that Q divides PS externally in the ratio 3:2.

i) Prove that Q has co-ordinates (-4ap, a(3-2p²)).​

ii) Show that as P varies, the locus of Q is another parabola. Find its equation and write down the co-ordinates of its vertex and focus in terms of a.​

​

The graph is attached. I need help with the second part. Thanks!

 

[KFunk]

If Q has the co-ordinates (-4ap, a(3-2p²)) then x= -4ap and y = a(3-2p²)

p = -x/4a so y=a(3 - 2x²/16a²)

y = 3a - x²/8a

x² = -8a(y - 3a)

Which is in the form (x - h)² = -4a(y - k)²

where the Vertex is at (h,k) and then the focus is 'a' units below the focus (or something like that )

 

[richz]

x=-4ap
p=x/-4a (1)

y=a(3-2p^2)
sub (1) into the above

y=a(3-2(x/-4a)^2)
y=3a-((2x^2)/16a)
16ay-48a^2 = -2x^2
16a(y-3a) = -2x^2

thats the equtation

so vertex is 0,3a

i think

 

[Dreamerish*~]

If Q has the co-ordinates (-4ap, a(3-2p2)) then x= -4ap and y = a(3-2p2)

p = -x/4a so y=a(3 - 2x2/16a2)

y = 3a - x2/8a

x2 = -8a(y - 3a)

Which is in the form (x - h)2 = -4a(y - k)2

where the Vertex is at (h,k) and then the focus is 'a' units below the focus (or something like that )

If you weren't twice as tall as me, I'd ask you to sit my maths trials for me.

Thanks!

 

[KFunk]

Well, if you can grow a few inches and sit my chem trial, then we've got something of a deal going on here.

 

[richz]

just wondering, which section of the paper would these sorts of questions be likely to be asked? like in q6-7 or q1-5?

between q 1-5 maybe even between 1-3

 

[Dreamerish*~]

Well, if you can grow a few inches and sit my chem trial, then we've got something of a deal going on here.

I don't think I can get you any more marks than you can get yourself in chem.

K3tan: this was from question 3.