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part ii is causing some problems .    

Marcop  Oct 27, 2014

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 #2
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+10

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
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6+0 Answers

 #1
avatar+92194 
+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

https://www.desmos.com/calculator/x2iznxi7bo

Melody  Oct 27, 2014
 #2
avatar
+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
avatar+92194 
+5

Thanks anon,

Have you followed this through to its conclusion?  Is it shorter?

Melody  Oct 27, 2014
 #4
avatar
+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
 #5
avatar+92194 
+5

Yes I expect that you are right but I think my method is okay too :)

Melody  Oct 28, 2014
 #6
avatar+83 
0

Thanks Melody and anonymous 

Marcop  Oct 28, 2014

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