#1**+10 **

I'll just do part ii) since you can already do part i)

If you start with the original parabola $${{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{ay}}$$

$$(2ap,ap^2)$$ This has a focal length of $$a$$ and a directrix of $${\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{a}}$$

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If you reflect this about the x axis it will turn upside down and you will have a parabola with a focal length of $${\left|{\mathtt{\,-\,}}{\mathtt{a}}\right|} = {\mathtt{a}}$$ and a directrix of $${\mathtt{y}} = {\mathtt{a}}$$ and the equation will be

The equation will be $$x^2=4(-a)y\;\;\rightarrow \;\;

x^2=-4ay$$

The parametric equation will be $$(2ap,-ap^2)$$

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If this is now translated down 2a units

The directix will translate down 2a units to $$y=a-2a\quad\rightarrow\quad y=-a$$

The focal length will still be $$a$$

The equation will be $$x^2=-4a(y+2a)$$

The parametric equation will be $$(2ap,-ap^2-2a)$$

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Therefore $$G(2ap,-ap^2-2a)$$

has the same focal length and directrix as the original parabola.

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Here is a graph to show what is happening

Melody Oct 27, 2014

#2**+10 **

Best Answer

An alternative approach, (to Melody's method), is to remove the parameter from the co-ordinates of G.

We have $$x=2ap,$$ from which $$p=x/2a$$, and on substitution, $$y=-2a-a(x/2a)^{2}$$, so $$x^{2}=-4ay-8a^{2}$$ etc..

Guest Oct 27, 2014

#3

#4**+5 **

Yes, I followed it to its conclusion. There isn't much to choose between them as far as length is concerned is there ? I do think though that my method is the one that the examiner expected (?).

Guest Oct 27, 2014