Locus and the Parabola (1 Viewer)

[atar90plus]

Hello
Just wondering if you guys can help me with this question. Please show full working out and solution preferably in latex if can

Consider the parabola (y-1)^2=8(x+2). Find

i) the coordinates of the vertex
ii) the coordinate of the parabola

 

[Fus Ro Dah]

Well the coordinate of the vertex is trivially (-2,1) very much like the centre of a circle. And I don't get what you mean by the coordinate of the parabola? I don't see a locus anywhere here...

 

[SpiralFlex]

Well the coordinate of the vertex is trivially (-2,1) very much like the centre of a circle. And I don't get what you mean by the coordinate of the parabola? I don't see a locus anywhere here...

It is a topic. This is not a locus question but is under its heading.

 

[gr_111]

ii) By coordinate of the parabola i presume you mean the focus and directrix.

The parabola is orientated along the x axis. The vertex, lies halfway between the focus and the directrix.

In your case , therefore a, or the focal length, is 2, therefore the focus is .

The directrix is equal to the focal length, but on the other side of the vertex. The vertex is already offset by -2 units so the directrix goes further again i.e. therefore equation of directrix is .

I don't feel I've explained it very clearly but thats the guts of it.